Solucao Calculo 5ed CAP.11

Texto

(1)Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers. (b) The terms a approach 8 as n becomes large. In fact, we can make a as close to 8 as we like by n. n. taking n sufficiently large. (c) The terms a become large as n becomes large. In fact, we can make a as large as we like by n. n. taking n sufficiently large. 2. (a) From Definition 1, a convergent sequence is a sequence for which lim a exists. Examples: n . { 1/n} ,. n. { 1/2n}. (b) A divergent sequence is a sequence for which lim a does not exist. Examples: { n} , { sin n} n n . n. 3. a =1 ( 0.2 ) , so the sequence is { 0.8,0.96,0.992,0.9984,0.99968,... } . n n+1 , so the sequence is 3n1. 4. a = n. 3 ( 1 ) 5. a = n n!. {. n. , so the sequence is. 2 3 4 5 6 , , , , ,... 2 5 8 11 14. {. }{ =. 3 3 3 3 3 , , , , ,... 1 2 6 24 120. 1,. 3 1 5 3 , , , ,... 5 2 11 7. }{ =. 3,. }. .. 3 1 1 1 , , , ,... 2 2 8 40. }. .. 6. a =2 4 6 ( 2n ) , so the sequence is n. { 2,2 4,2 4 6,2 4 6 8,2 4 6 8 10,... } ={ 2,8,48,384,3840,... } .. 7. a =3 , a 1. =2a 1 . Each term is defined in terms of the preceding term.. n+1. n. a =2a 1=2(3)1=5 . a =2a 1=2(5)1=9 . a =2a 1=2(9)1=17 . a =2a 1=2(17)1=33 . The 2. 1. 3. 2. 4. 3. 5. 4. sequence is { 3,5,9,17,33,... } . a 8. a =4 , a 1. =. n+1. n. a 1. . Each term is defined in terms of the preceding term.. n. a. a 2 4 4 4/3 4/3 a= = = .a = = = =4 . Since a =a , we can see that the terms of the 2 a 1 41 3 3 a 1 3 1 4 1/3 1 2 1 3 4 4 sequence will alternately equal 4 and 4/3 , so the sequence is 4, ,4, ,4,... . 3 3 1. {. }. 1.

(2) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 9. The numerators are all 1 and the denominators are powers of 2 , so a = n. 1 n. .. 2. 10. The numerators are all 1 and the denominators are multiples of 2 , so a = n. 1 . 2n. 11. { 2,7,12,17,... } . Each term is larger than the preceding one by 5 , so a =a +d(n1)=2+5(n1)=5n3 . n. 1. 12.. {. . 1 2 3 4 , , , ,... 4 9 16 25. }. 2. . The numerator of the n th term is n and its denominator is ( n+1 ) .. Including the alternating signs, we get a = ( 1 ). 13.. {. 2 4 8 , , ,... 3 9 27. 1,. }. n. n. n. ( n+1 ). . Each term is . 2. .. 2 2 times the preceding one, so a = n 3 3. n1. .. 14. { 5,1,5,1,5,1,... } . The average of 5 and 1 is 3 , so we can think of the sequence as alternately n+1. adding 2 and 2 to 3 . Thus, a =3+(1) n. 2.. 15. a =n(n1) . a as n , so the sequence diverges. n. 16. a = n. n. n+1 1+1/n 1+0 1 = = as n . Converges , so a n 3n1 31/n 30 3 2. 17. a = n. 18. a = n. 3+5n. 2. n+n n. 1+ n n. 19. a = n. 20. a = n. 2. n+1. 3. =. 2 2 2 ( 5+3/n 3+5n ) /n = = 2 2 ( n+n ) /n 1+1/n. =. 1 3. , so a n. 5+0 =5 as n . Converges 1+0. 1 1 =1 as n . Converges , so a n 0+1 1/ n +1 2 3. n. , so lim a = n . n. 1 lim 3 n . 2 3. n. =. 1 2 0=0 by (8) with r= . Converges 3 3. n n = . The numerator approaches and the denominator approaches 0+1=1 as 1+ n 1/ n +1 2.

(3) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. n , so a as n and the sequence diverges. n. 21. a =. n1. ( 1 ). n. ( 1 ). =. n1. , so 0 a =. 2 n+1/n n +1 Theorem and Theorem 6. Converges n. n 3. (1) n. n. n. 1 1 0 as n , so a 0 by the Squeeze n n+1/n n. 3. 1 1 as n , but the terms of the sequence 3 2 3 2 n n 2 1 n +2n +1 n +2n +1 1+ + 3 n n {a } alternate in sign, so the sequence a ,a ,a ... converges to 1 and the sequence a ,a ,a ,.... 22. a =. . Now a =. =. 1 3 5,. n. 2 4 6. converges to +1 . This shows that the given sequence diverges since its terms don’t approach a single real number. 23. a =cos ( n/2) . This sequence diverges since the terms don’t approach any particular real number n. as n . The terms take on values between 1 and 1 . 24. a =cos (2/n) . As n , 2/n 0 , so cos (2/n) cos 0=1 . Converges n. 25. a = n. (2n1)! (2n1)! 1 = = 0 as n . Converges (2n+1)! (2n+1)(2n)(2n1)! (2n+1)(2n). 26. 2n as n , so since lim arctanx= x . n. 27. a = n. 28. a = n. n. n. e +e. . 2n. e 1. e. n. e. 2n. =. 1+e n. n. e e. . 1+0 n. e 0. , we have lim arctan2n= . Converges 2 2 n . 0 as n . Converges. ln n ln n 1 1 = = 1 as n . Converges ln 2n ln 2+ln n ln 2 0+1 +1 ln n 2 n. 29. a =n e = n. 2. n e. n. 2. . Since lim. x . x e. x. =lim. x . 2x e. x. =lim. x . 2 e. x. =0 , it follows from Theorem 3 that lim a =0 . n . n. Converges 3.

(4) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. n. 30. a =ncos n =n(1) . Since a =n as n , the given sequence diverges. n. n. 2. 31. 0. cos n n. . 1. , so since lim. n. n . 2. 2. n+1 n. 32. a =ln (n+1)ln n=ln n. 33. a =nsin (1/n)= n. 1 n. { } 2. =0 ,. 2. =ln. cos n n. converges to 0 by the Squeeze Theorem.. 2. 1+. 1 n. ln (1)=0 as n . Converges. sin (1/x) sin t sin (1/n) . Since lim =lim [] where t=1/x]=1 , it follows from 1/x t 1/n + x t 0. Theorem 3 that. {a } n. 2. 34. a = n n 1 = n. converges to 1. 1 n. 2. n . 2. n. 1. 1 2. =n. n. 1 n. 1. 1 2. n(01) n as n ,. n. so a as n . Diverges n. 2 35. a = 1+ n n. 1/n. ln a = n. 1 ln n. 1+. 2 n. . As n ,. 1 0 and ln n. 1+. 2 n. 0 , so ln a 0 . n. 0. Thus, a e =1 as n . Converges n. 1 sin 2n 1 1 1 . a and lim =0 , so a lim a =0 by the n 1+ n n n 1+ n n 1+ n 1+ n n 1+ n n Squeeze Theorem. Converges 36. a =. 37. { 0,1,0,0,1,0,0,0,1,... } diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to either one (or any other value) for n sufficiently large.. {. }. 1 1 1 1 1 1 1 1 1 1 . a = and a = for all positive integers n . lim a =0 , , , , , , , ,... 2n1 2n n n n+2 1 3 2 4 3 5 4 6 n 1 1 since lim a =lim =0 and lim a =lim =0 . For n sufficiently large, a can be made as 2n1 2n n n n n n n n+2 close to 0 as we like. Converges 2 3 ( n1 ) n 1 n n n! 1 39. a = = diverges. = as n , so a n n n 2 2 2 2 2 2 2 4 2 38.. { }. 40. 4.

(5) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. n. 3 3 3 3 3 3 3 3 3 27 0< a = = = 0 as n , so by the n n! 1 2 3 (n1) n 1 2 n 2n Squeeze Theorem and Theorem 6,. { ( 3) n/n}. converges to 0 .. 41.. From the graph, we see that the sequence and 1 (approximately).. {(. 1 ). n. n+1 n. }. is divergent, since it oscillates between 1. 42.. From the graph, it appears that the sequence converges to 2 . 2 2 n converges to 0 by (6), and hence 2+ . {. }. {. }. n. converges to 2+0=2 .. 43.. From the graph, it appears that the sequence converges to about 0.78 . 2n 2 2n lim =lim =1 , so lim arctan =arctan1= . 2n+1 4 n 2n+1 n 2+1/n n 5.

(6) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 44.. From the graph, it appears that the sequence converges (slowly) to 0 . sin n 1 0 0 as n , so by the Squeeze Theorem and Theorem 6, n n to 0 .. { } sin n n. converges. 45.. From the graph, it appears that the sequence converges to 0 . 3. 0<a = n. n n n n 1 1 1 1 = n! n (n1) 3 2 1 ( n2 ) ( n3) 2. n [for n 4] (n1)(n2)(n3) 1/n = 0 as n ( 11/n ) ( 12/n ) ( 13/n ) So by the Squeeze Theorem,. { n3/n!}. converges to 0 .. 46.. From the graph, it appears that the sequence converges to 5. 6.

(7) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. n. 5=. n 5 . =. n n. n. n. 3 +5 . n. n. n. n n. 5 +5 = 2. n. 5 1/n. 0. 2 5 5asn lim 2 =2 =1] n . Hence, a 5 by the Squeeze Theorem. n. ( x x) 1/x. Then x x x x ln ( 3 +5 ) 3 ln 3+5 ln 5 =lim. Alternate Solution: Let y= 3 +5 lim ln y = lim. x . x . = lim. x . so lim y=e x . ln 5. x. x. x . x. 3 +5. x. 3 5. ln 3+ln 5. =5 , and so. =ln 5. x. 3 5. +1. {. n. n. n. 3 +5. }. converges to 5 .. 47.. From the graph, it appears that the sequence approaches 0 . 1 3 5 2n1 1 3 5 ( 2n1 ) = 0<a = n n 2n 2n 2n 2n ( 2n ) 1 1 (1) (1) (1)= 0 as n 2n 2n 1 3 5 ( 2n1 ) So by the Squeeze Theorem, converges to 0 . n ( 2n ). {. }. 48.. 7.

(8) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. From the graphs, it seems that the sequence diverges. a = n. 1 3 5 ( 2n1 ) . We first n!. n1. 3 for all n . This is clearly true for n=1 , so let P(n) be the n 2 statement that the above is true for n . We must show it is then true for n+1 . 2n+1 3 2n+1 3 n1 2n+1. . a =a (induction hypothesis). But , and so we get that n+1 n n+1 n+1 2 n+1 2 3 n1 3 3 n a = which is P(n+1) . Thus, we have proved our first assertion, so since n+1 2 2 2 3 n1 diverges (by (8)), so does the given sequence a . n 2 prove by induction that a . {. }. { }. n. 49. (a) a =1000(1.06) a =1060 , a =1123.60 , a =1191.02 , a =1262.48 , and a =1338.23 . 1. n. 2. 3. 4. 5. n. (b) lim a =1000lim (1.06) , so the sequence diverges by (8) with r=1.06>1 . n . 50. a. =. n+1. n. {. n . 1 if a is an even number a n 2 n 3a +1 if a is an odd number n. When a =11 , the first 40 terms are 11 , 34 , 17 1. n. , 52 , 26 , 13 , 40 , 20 , 10 , 5 , 16 , 8 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 . When a =25 , the first 40 terms are 25 , 76 , 38 , 19 , 58 , 29 , 88 , 44 , 22 1. , 11 , 34 , 17 , 52 , 26 , 13 , 40 , 20 , 10 , 5 , 16 , 8 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 , 2 , 1 , 4 . The famous Collatz conjecture is that this sequence always reaches 1 , regardless of the 8.

(9) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. starting point a . 1. 51. If r 1 , then. { r n}. diverges by (8), so x. x. r <1 then lim xr =lim x . x. x . x . ( ln r ) r. n. n. diverges also, since nr =n r r x. 1. =lim. r converges whenever r <1 .. { nr n}. x. =lim. x . r n =0 , so lim nr =0 , and hence ln r n . n. . If. { nr n}. 52. (a) Let lim a =L . By Definition 1, this means that for every >0 there is an integer N such that n . n. a L < whenever n>N . Thus, a n. lim a. n . =L and so lim a =lim a. n+1. n . n. (b) If L=lim a then lim a n. n . n . n . L < whenever n+1>N n>N1 . It follows that. n+1. n+1. . 2. =L also, so L must satisfy L=1/ ( 1+L ) L +L1=0 L=. n+1. 1+ 5 2. (since L has to be nonnegative if it exists).. { a } is a decreasing sequence, a >a for all n 1 . Because all of its terms lie between 5 and 8 , { a } is a bounded sequence. By the Monotonic Sequence Theorem, { a } is convergent; that is, { a } has a limit L . L must be less than 8 since { a } is decreasing, so 53. Since. n. n. n+1. n. n. n. n. 5 L<8 . n. 54. a =1/5 defines a decreasing geometric sequence since a n. =. n+1. 1 a <a for each n 1 . The 5 n n. 1 sequence is bounded since 0<a for all n 1 . n 5 1 1 1 1 is decreasing since a = = < =a for each n 1 . The sequence is n 2n+3 n+1 2(n+1)+3 2n+5 2n+3 n 1 1 bounded since 0<a for all n 1 . Note that a = . n 5 1 5 55. a =. 2n3 2x3 defines an increasing sequence since for f (x)= , n 3n+4 3x+4 1 / (3x+4)(2)(2x3)(3) 17 f (x)= = >0 . The sequence is bounded since a a = for n 1 , n 1 2 2 7 ( 3x+4 ) ( 3x+4 ) 2n3 2n 2 and a < < = for n 1 . n 3n 3n 3 56. a =. 9.

(10) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 57. a =cos ( n /2 ) is not monotonic. The first few terms are 0 , 1 , 0 , 1 , 0 , 1 , 0 , 1 , ... . In fact, n the sequence consists of the terms 0 , 1 , 0 , 1 repeated over and over again in that order. The sequence is bounded since a 1 for all n 1 . n. n. x. 58. a =ne. defines a positive decreasing sequence since the function f (x)=xe is decreasing for x>1. n. x. /. x. x. 1 and below by 0. e. . [ f (x)=e xe =e (1x)<0] for x>1 .] The sequence is bounded above by a = 1. n. 59. a = n. /. f. x. defines a decreasing sequence since for f (x)=. 2. n +1. ,. 2. x +1. 2 2 ( 1 x +1 ) ( 1 ) x(2x) 1x (x)= = 0 for x 1 . The sequence is bounded since 0<a n 2 ( x2+1) 2 ( x2+1) 2. for all. n 1 . 60. a =n+ n. 1 1 defines an increasing sequence since the function g(x)=x+ is increasing for x>1 . n x. /. 2. [g (x)=11/x >0] for x>1 .] The sequence is unbounded since a as n . (It is, however, n. bounded below by a =2 .) 1. 1/2. 61. a =2 1. 3/4. , a =2 2. 7/8. , a =2 3. , .... n n n ( 2 1 ) /2 1 ( 1/2 ) =2 , so a =2. ( n) =21=2 .. 1 1/2. . lim a =lim 2. n. n . n. n . Alternate solution : Let L=lim a . (We could show the limit exists by showing that n . n. {a }. is. n. 2. bounded and increasing.) Then L must satisfy L= 2 L L =2L L(L2)=0 . L 0 since the sequence increases, so L=2 . 62. (a) Let P be the statement that a n. a and a 3 . P is obviously true. We will assume that P. n+1. n. is true and then show that as a consequence P. 2+a. 2+a a. n+1. n. n. n+1. 1. must also be true. a. a , which is the induction hypothesis. a. n+1. n. a. n+2. 3. n+1. n+1. 2+a. n+1. 2+a. n. n. 2+a 3 2+a 9. n. n. a 7 , which is certainly true because we are assuming that a 3 . So P is true for all n , and so n. n. n. a a 3 (showing that the sequence is bounded), and hence by the Monotonic Sequence Theorem, 1. n. lim a exists.. n . n. (b) If 10.

(11) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. L=lim a , then lim a n. n . n . 2. 2. =L also, so L= 2+L L =2+L L L2=0 (L+1)(L2)=0 L=2 (since. n+1. L can’t be negative).. {a }. 63. We show by induction that Let P be the proposition that a n. a. >a . n+1. n. Now a. 1 a. =3. n. 1 a. n. 1. n. n+1. n+1. n. n. 1 >3 =a P . This proves that n+1 n+1 a. {a } n. is increasing and bounded above by. n. 3 , so 1=a <a <3 , that is, 1. >a and 0<a <3 . Clearly P is true. Assume that P is true. Then. n+1. 1 1 1 . < > a a a. n+1. n+2. is increasing and bounded above by 3 .. n. n. {a }. is bounded, and hence convergent by the Monotonic Sequence. n. Theorem. If L=lim a , then lim a n. n . , so L=. n . 2. =L also, so L must satisfy L=31/L L 3L+1=0 L=. n+1. 3+ 5 . 2. 64. We use induction. Let P be the statement that 0<a n. a =1/(32)=1 . Now assume that P is true. Then a 2. a. n. =. n+2. 3

(12) 5 . But L>1 2. 1 3a. . n+1. a 2 . Clearly P is true, since. n+1. 1. n. a a. n+1. n. a 3a. n+1. n. 3a . n+1. n. 1 . Also a >0 (since 3a is positive) and a 2 by the induction =a n+1 n+2 n+1 n+1 3a n. hypothesis, so P. n+1. is true.. To find the limit, we use the fact that lim a =lim a n . L 2 , so we must have L=. n. n . n+1. L=. 3

(13) 5 1 2 . But L 3L+1=0 L= 2 3 L. 3 5 . 2. 65. (a) Let a be the number of rabbit pairs in the n th month. Clearly a =1=a . In the n th month, 1. n. each pair that is 2 or more months old (that is, a pairs already present. Thus, a =a n. f (b) a = n. n+1. f. n. a. f =. n1. f. n. n1. f =. +a. n1. +f. n1. f. n1. n2. n2. n2. , so that f. =1+. f. n2 n1. 2. pairs) will produce a new pair to add to the a. {a } ={ f } n. =1+. n. f. n1. 1 f. /. , the Fibonacci sequence. =1+. n2. n1. 1 a. n2. . If L=lim a , then n . n. 11.

(14) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. L=lim a n . n1. and L=lim a n . n2. , so L must satisfy L=1+. 1+ 5 1 2 L L1=0 L= (since L must be 2 L. positive). 66. (a) If f is continuous, then f (L)= f lim a n . n. ( ). =lim f a =lim a n . n. n . =L by Exercise 52(a).. n+1. (b) By repeatedly pressing the cosine key on the calculator (that is, taking cosine of the previous answer) until the displayed value stabilizes, we see that L 0.73909 .. 67. (a). { } 5. From the graph, it appears that the sequence. n n!. 5. n =0 . converges to 0 , that is, lim n n!. (b). From the first graph, it seems that the smallest possible value of N corresponding to =0.1 is 9 , since 5. 5. n /n!<0.1 whenever n 10 , but 9 /9!>0.1 . From the second graph, it seems that for =0.001 , the smallest possible value for N is 11 . 12.

(15) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 68. Let >0 and let N be any positive integer larger than ln ( ) /ln r . If n>N then n>ln ( ) /ln r nln r <ln ln 69. If lim. n . ( r n) <ln . n. n. n. r < r 0 < , and so by Definition 1, lim r =0 . n . a =0 then lim a =0 , and since a a a n. n. n . n. n. n. , we have that lim a =0 by the n . n. Squeeze Theorem. 70. (a) b. n+1. n+1. a ba. n. n1. n. n1. =b +b <b +b. a+b b+b. n2 2. n3 3. n1. n. n2 2. n3 3. n1. n. a +b. b +b. (b) Since ba>0 , we have b .. n+1. a + +ba. a. b + +bb. n+1. n. +a. +b =(n+1)b. <(n+1)b (ba) b n. n+1. (c) With this substitution, (n+1)anb=1 , and so b = 1+. (e) a <a. 2n. n. (f) Since. since. {a }. {a } n. n. (n+1)b (ba)<a. 1 n. n 1 1 <1 2 2n is increasing, so a <a <4 .. 1+. (d) With this substitution, we get. n. n. n. <a 1+. n+1. n+1. 1 2n. = 1+. 1 n+1. n. <2. 1+. n+1. n+1. . 1 2n. 2n. <4 .. 2n. is increasing and bounded above by 4 , a a 4 , and so 1. n. n. b (n+1)anb <a. n. {a } n. is bounded and. monotonic, and hence has a limit by Theorem 11. 71. (a) First we show that a>a >b >b . 1. 1. 2 a+b 1 1 ab = ( a2 ab +b ) = ( a b ) >0 (since a>b ) a >b . Also 1 1 2 2 2 1 1 aa =a (a+b)= (ab)>0 and bb =b ab = b ( b a ) <0 , so a>a >b >b . In the same way we 1 1 1 1 2 2 can show that a >a >b >b and so the given assertion is true for n=1 . Suppose it is true for n=k , that. a b = 1. 1. 1. is, a >a. >b. b. =. k. a. k+2. a. k+1. k+2. a. k+1. k+2. 2. 2. 1. >b . Then. k+1. k. 1 1 a +b a b = a 2 a b +b k+1 k+1 2 k+1 k+1 k+1 k+1 2 k+1 k+1 2 1 = a b >0 k+1 k+1 2 1 1 = a (a +b )= a b >0 k+1 2 k+1 k+1 2 k+1 k+1. (. ). (. ). 13.

(16) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. b. and b. k+1. . =b. k+2. k+1. a. b. k+1 k+1. = b. b. k+1. k+1. . <0 a. a. k+1. >a. k+1. >b. k+2. >b. k+2. k+1. assertion is true for n=k+1 . Thus, it is true for all n by mathematical induction. (b) From part (a) we have a>a >a >b >b >b , which shows that both sequences, n. {b }. n+1. n+1. n. , so the. {a } n. and. , are monotonic and bounded. So they are both convergent by the Monotonic Sequence. n. Theorem. a +b (c) Let lim a = and lim b = . Then lim a n . n. n . 2 = + = .. n. n . n. =lim. n+1. n. =. 2. n . + 2. 72. (a) Let >0 . Since lim a =L , there exists N such that a L < for n>N . Since 2n. n . 2n+1. 2. 2n. 1. {. }. L < for n>N . Let N=max 2N ,2N +1. =L , there exists N such that a. lim a. n . 1. 2n+1. 2. 1. 2. and let. n>N . If n is even, then n=2m where m>N , so a L = a L < . If n is odd, then n=2m+1 , 1. where m>N , so a L = a 2. 2m. n. L < . Therefore lim a =L .. 2m+1. n. n . n. 1 3 1 7 1 17 = =1.5 , a =1+ = =1.4 , a =1+ = =1.416 , 1 2 3 4 1+1 2 5/2 5 12/5 12 1 41 1 99 1 239 a =1+ = 1.413793 , a =1+ = 1.414286 , a =1+ = 1.414201 , 5 6 7 29/12 29 70/29 70 169/70 169 1 577 a =1+ = 1.414216 . Notice that a <a <a <a and a >a >a >a . It appears that the odd 8 1 3 5 7 2 4 6 8 408/169 408 terms are increasing and the even terms are decreasing. Let’s prove that a >a and a <a by (b) a =1 , a =1+. 2n2. mathematical induction. Suppose that a. >a. 2k2. 2k. . Then 1+a. >1+a . 2k2. 2k. 2n. 1 1+a. <. 2k2. 1+. 1 1+a. <1+. 1 1+a. >1+. 2k2. 1+. 2k1. 1 1+a. a. <a. 2k1. 2k+1. 2k. 1 1+a. 1+a. <1+a. 2k1. 2k+1. . 1 1+a. 2k1. a >a 2k. 2k+2. >. 1 1+a. 2n1. 1 1+a. n. . 2k. . 2k+1. . We have thus shown, by induction, that the odd terms are. 2k+1. increasing and the even terms are decreasing. Also all terms lie between 1 and 2 , so both. {b }. 2n+1. {a } n. and. are bounded monotonic sequences and are therefore convergent by Theorem 11. Let. 4+3a 1 1 n , lim a =L . Then lim a =L also. We have a =1+ =1+ = 2n 2n+2 n+2 1+1+1/(1+a ) (3+2a )/(1+a ) 3+2a n n n n n n. so 14.

(17) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.1 Sequences. 4+3a. 2n. =. a. 2n+2. . Taking limits of both sides, we get L=. 3+2a. 2n. 4+3L 2 2 3L+2L =4+3L L =2 L= 2 3+2L. (since L>0 ). Thus, lim a = 2 . Similarly we find that lim a 2n. n . {p }. 73. (a) Suppose n. =. n+1. a+ p. lim p n . n. = 2 . So, by part (a), lim a = 2 .. 2n+1. n . n. converges to p . Then blim p. bp p. n. n . =. n+1. n. n . a+lim p. p=. n. n . bp 2 p +ap=bp p( p+ab)=0 p=0 or p=ba . a+ p. b p p n a n b n = < p since 1+ >1 . (b) p = n+1 a+ p a a n p n n 1+ a b b b 2 p ,p< p< p ,p< (c) By part (b), p < 1 0 2 1 0 3 a a a n n b b p< p , so lim p lim p =0 since b<a . n 0 n 0 a a n n bp. b a. 3. b a. p< 2. p , etc. In general, 0. (d) Let a<b . We first show, by induction, that if p <ba , then p <ba and p 0. For n=0 , we have p p = 1. 0. bp. 0. p =. 0. 0. a+ p. 0. 0. 1. =ba. ba p. k+1. a(ba)+bp ap bp. a+ p. k. =. p. p. k+2. bp =. k+1. k+1. a+ p. k. k. a+ p. k. k. p. p =. k+1. > p . Then. k+1. k. a(ba p ) =. k. >0 because p <ba . So p k. a+ p. <ba . And. k+1. k. (ba p. k+1. k+1. 0. 0 k. k. n. >0 since p <ba . So p > p .. Now we suppose the assertion is true for n=k , that is, p <ba and p bp. >p .. n+1. p (ba p ). 0. a+ p. n. ). k+1. >0 since p. <ba . Therefore, p. k+1. a+ p. >p. k+2. k+1. . Thus, the. k+1. assertion is true for n=k+1 . It is therefore true for all n by mathematical induction. A similar proof by induction shows that if p >ba , then p >ba and p is decreasing. In either case the sequence 0. {p } n. n. { } n. is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem. It then. follows from part (a) that lim p =ba . n . n. 15.

(18) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 1. Using Theorem 5 with . . n. b (x5) , b =. n=0 n. n. f. ( n). ( 8) (a) f (5) , so b = . 8 n! 8! /. 2. (a) Using Formula 6, a power series expansion of f at 1 must have the form f (1)+ f (1)(x1)+ /. . Comparing to the given series, 1.60.8(x1)+ , we must have f (1)=0.8 . But from the graph, /. f (1) is positive. Hence, the given series is not the Taylor series of f centered at 1 . 1 / / / 2 (b) A power series expansion of f at 2 must have the form f (2)+ f (2)(x2)+ f (2)(x2) + . 2 1 / / 2 3 f (2)=1.5 Comparing to the given series, 2.8+0.5(x2)+1.5(x2) 0.1(x2) + , we must have 2 / /. ; that is, f (2) is positive. But from the graph, f is concave downward near x=2 , so f be negative. Hence, the given series is not the Taylor series of f centered at 2 .. / /. (2) must. 3. ( n) ( n) f ( 0) f (x) 0 cos x 1 1 sin x 0 2 cos x 1 3 sin x 0 4 cos x 1 We use Equation 7 with f (x)=cos x .. n. cos x = f (0)+ f /(0)x+ f 2. ( 3) ( 4) (0) 2 f (0) 3 f ( 0 ) 4 x+ x+ x + 2! 3! 4!. / /. n 2n. 4. ( 1 ) x x x =1 + = n=0 (2n)! 2! 4! n 2n. ( 1 ) x If a = n ( 2n )! a lim. n . n+1. a. n. , then. 2n+2. =lim. n . x (2n)! 2n (2n+2)! x. 2. =x lim. n . 1. ( 2n+2 ) ( 2n+1 ). =0<1 for all x . 1.

(19) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. So R= (Ratio Test). 4. n 0 1 2. ( n) f (x) sin 2x 2cos 2x. 3. 2 cos 2x. 4. 2 sin 2x . f 0 2. 2. 0. 3. 2. 2 sin 2x 4. . ( n). ( 0). 3. 0 . n 2n+1 ( 2n+1 ) , so (0)=0 if n is even and f (0)=(1) 2 ( n) ( 2n+1 ) (0) 2n+1 sin 2x = f (0) xn= f x n=0 n=0 (2n+1)! n!. f. ( n). n 2n+1 2n+1. (1) 2 x = n=0 (2n+1)! 2 2 a 2 x n+1 =lim =0<1 for all x , a n (2n+3)(2n+2) . lim. n . n. so R= (Ratio Test). 5. ( n). n. f. 0. ( 1+x ). 1. 3(1+x). 2. 12(1+x). 3. 60(1+x). 4. 360(1+x) . . f. (x) 3. ( n). ( 0). 1. 4 5 6 7. 3 12 60 360 . 2.

(20) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. ( 1+x ). 3. / / / ( 4) (0) 2 f (0) 3 f (0) 4 = f (0)+ f (0)x+ x+ x+ x + 2! 3! 4! 4 3 2 5 4 3 3 6 5 4 3 4 x x+ x =13x+ 3! 4! 2! 4 3 2 2 5 4 3 2 3 6 5 4 3 2 4 =13x+ x x+ x 2 2! 2 3! 2 4!. f. /. n. / /. n. n. (1) (n+2)(n+1)x (1) (n+2)! x = = n=0 n=0 2(n!) 2 . a lim. n . n+1. a. (n+3)(n+2)x 2. =lim. n . n. n+1. n. 2. . (n+2)(n+1)x. n. = x lim. n . n+3 = x <1 for convergence, n+1. so R=1 (Ratio Test). 6. n 0 1. ( n) f (x) ln (1+x). f 0. 1. 1. 2. ( 1+x ). 3. 2(1+x). 4. 6(1+x). 5. 24(1+x) . . ( 1+x ). 2. ( n). ( 0). 1. 3. 2. 4. 6. 5. 24 /. ln (1+x) = f (0)+ f (0)x+. f. / /. / / /. (0) 2 f (0) 3 x+ x 2! 3!. ( 5) (0) 4 f (0) 5 + x+ x + 4! 5! 1 2 2 3 6 4 24 5 =x x + x x+ x 2 6 24 120. f. ( 4). 2. 3. 4. 5. ( 1 ) x x x x =x + + = n=1 2 3 4 5 n. n1. x. n. 3.

(21) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. a lim. n+1. a. n . n+1. x n n+1 n x. =lim. n. n . (x). f. =lim. n . x = x <1 for convergence, so R=1 . 1+1/n. 7. n. f. ( n) 5x. 0. e. 1. 5e. 2. 5e. 3. 5e. 4. 5e . . e =. ( 0). 1. 5x. 5. 2 5x. 25. 3 5x. 125. 4 5x. 625 n. (n). f (0) n 5 n x = x . n=0 n! n!. . 5x. ( n). n=0. a lim. n+1. a. n . n+1. 5. = lim. n . n. n+1. x (n+1)!. . n! n. 5 x. n. 5 x = lim n+1 =0<1forallx,soR= . n 8. n. f. ( n). (x). x. 0. xe. 1. (x+1)e. 2. (x+2)e. 3. (x+3)e . x. xe =. n=0. f. ( n). ( 0). 0 x. 1. x. 2. x. 3 . (n). n. f (0) n n n n n x . x = x = x = n=0 n! n=1 n! n=1 (n1)! n! 4.

(22) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. a lim. a. n . n+1. n+1. x n!. =lim. n . n. . (n1)! x. n. =lim. n . x =0<1 for all x , so R= . n. 9. ( n) f (x) sinh x cosh x sinh x cosh x sinh x . n 0 1 2 3 4 f. ( n). (0)=. {. f 0 1 0 1 0 . ( n). ( 0). 0 if n is even 1 if n is odd. so sinh x=. n=0. 2n+1. x . ( 2n+1 )!. 2n+1. x Use the Ratio Test to find R . If a = , then n ( 2n+1 )! 2n+3 a x n+1 (2n+1)! lim = lim 2n+1 (2n+3)! a n n x n 1 2 =0<1 = x lim n (2n+3)(2n+2) for all x , so R= . 10. ( n) f (x) cosh x sinh x cosh x sinh x . n 0 1 2 3 . {. f 1 0 1 0 . ( n). ( 0). 2n. x 1 if n is even so cosh x= . f (0)= n=0 ( 2n )! 0 if n is odd Use the Ratio Test to find R . If. ( n). 5.

(23) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 2n. x a= , then n (2n)! a n+1 lim = lim a n n n. 2n+2. x (2n)! 2n (2n+2)! x 1 2 = x lim (2n+2)(2n+1) =0<1 n . for all x , so R= . 11. ( n) f ( 0) 1 5 2 2 3 0 4 0 0 2 2 n f (x) =7+5(x2)+ (x2) + (x2) n=3 n! 2!. n. ( n) f (x) 1+2x 2 0 0 . =7+5(x2)+ ( x2 ). 2. Since a =0 for large n , R= . n. 12. n. f. ( n) 3. f. ( n). ( 0). 1. 0. x. 1. 3x 6x 6 0 0 . 2 3 4 5 . (x). 2. 3 6 6 0 0 6.

(24) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. f (x) =1+3(x+1). 6 2 6 3 (x+1) + (x+1) 3! 2! 2. =1+3(x+1)3(x+1) + ( x+1 ). 3. Since a =0 for large n , R= . n. 13. Clearly, f a lim. x. (x)=e , so f. ( n). 3. n+1. a. n . ( n). =lim. n . n. x. (3)=e and e = 3. n+1. e (x3) (n+1)!. n=0. n! 3. 3. . n. 3. e e n n (x3) . If a = (x3) , then n n! n! x3 =0<1 for all x , so R= . n+1. =lim. n . e (x3). 14. 0. ( n) f (x) ln x. 1. x. 2. x. 3. 2x. 4. 3 2x. . . n. f. ( n). ( n) f ( 0) ln 2 1 2 1 4 2 8 3 2 16 . 1 2 3 4. n1. (2)=. (1). (n1)!. n. for n 1 , so ln x=ln 2+. 2 a lim. n . n+1. a. n. =. n=1. n1. (1). n. (x2) n. .. n 2. n x2 x2 lim = <1 for convergence, so x2 <2 R=2 . 2 n n+1 2. 15. n 0. ( n) f (x) cos x. ( n) f ( 0) 1 7.

(25) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 1 2 3 4 . sin x cos x sin x cos x . cos x= lim. k=0. 2. 4. 6. 2n. f ( ) (x ) (x ) (x ) k n+1 (x ) (x ) =1+ + = (1) n=0 2! 4! 6! (2n)! k!. n+1. a. n . (k). . a. 0 1 0 1 . 2n+2. x (2n)! 2n (2n+2)! x. =lim. n . n. .. 2. x =lim =0<1 for all x , so R= . n (2n+2)(2n+1). 16. n 0 1 2 3 4 . ( n) f (x) sin x cos x sin x cos x sin x . ( n) f ( 0) 1 0 1 0 1 (k). sin x = f ( /2) k=0 k!. x. 2. 2. k. 4. 6. (x /2) (x /2) (x /2) =1 + + 2! 4! 6! . 2n. (x /2) = (1) n=0 (2n)! a lim. n+1. a. n . n. n. 2n+2. x /2 (2n+2)!. =lim. n . . 2. (2n)! x /2. 2n. x /2 =lim =0<1 for all x , so R= . n (2n+2)(2n+1). 17. n. f. ( n). (x). f. ( n). ( 0) 8.

(26) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 0. x. 1. . 1 3 1 1 2 3 3 1 3 2 2. 1/2. 1 3/2 x 2. 3 5/2 x 4. 2. . 3 . 15 7/2 x 8. . . 1 2. . 3 2. 1. . 5. 3 . . 5 2. . 1 7. 3. . 2. 3. 1 1 1 3 (x9) 3 5 (x9) = (x9)+ + 3 2 5 3 7 2! 3! 3 x 2 3 2 3 2 3 n 1 3 5 (2n1) n (x9) = (1) . n 2n+1 n=0 2 3 n!. a lim. n+1. a. n . n. 1 3 5 (2n1)[2(n+1)1] x9. =lim. n+1 [2(n+1)+1]. n . 2. (2n+1) x9. =lim. 2. n . 2 3 (n+1). 3. (n+1)!. =. n+1. n. . 2n+1. 2 3. !. 1 3 5 (2n1) x9. n. 1 x9 <1 9. for convergence, so x9 <9 and R=9 . 18. ( n). n. f. 0. x. 1. 2x. 2. 6x. (x). 2. f. ( n). ( 0). 1 3. 4. 2 6 9.

(27) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 5. 3. 24x. 4. 120x . x. 2. 24. 6. 120 2. 3. 4. (x1) (x1) (x1) =12(x1)+6 24 +120 2! 3! 4! 2. 3. 4. =12(x1)+3(x1) 4(x1) +5(x1) =. n=0. a lim. n+1. a. n . n. n. (1) (n+1)(x1) .. =lim. (n+2) x1. n+1 n. n . =lim. n . n+2 x1 n+1. = x1 <1 for convergence, so R=1 .. (n+1) x1 ( n+1 ) ( n+1 ) (x)= sin x or cos x . In each case, f (x) 1 , so by Formula 9 19. If f (x)=cos x , then f 1 n+1 x with a=0 and M=1 , R (x) . Thus, R (x) 0 as n by Equation 10. So n n (n+1)! lim R (x)=0 and, by Theorem 8, the series in Exercise 3 represents cos x for all x . n . n. n. ( n+1 ) (x)= sin x or cos x . In each case, f (x) 1 , so by Formula 9 n+1 1 x with a=0 and M=1 , R (x) . Thus, R (x) 0 as n by Equation 10. So n n (n+1)! 2 lim R (x)=0 and, by Theorem 8, the series in Exercise 16 represents sin x for all x .. 20. If f (x)=sin x , then f. n . ( n+1 ). n. ( n+1 ) (x)=cosh x or sinh x . Since sinh x < cosh x =cosh x for all 21. If f (x)=sinh x , then for all n , f ( n+1 ) x , we have f (x) cosh x for all n . If d is any positive number and x d , then ( n+1 ) f (x) cosh x cosh d , so by Formula 9 with a=0 and M=cosh d , we have cosh d n+1 R (x) x . It follows that R (x) 0 as n for x d (by Equation 10). But d n n ( n+1 )! was an arbitrary positive number. So by Theorem 8, the series represents sinh x for all x . ( n+1 ) (x)=cosh x or sinh x . Since sinh x < cosh x =cosh x for all 22. If f (x)=cosh x , then for all n , f ( n+1 ) x , we have f (x) cosh x for all n . If d is any positive number and x d , then 10.

(28) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. ( n+1 ). (x) cosh x cosh d , so by Formula 9 with a=0 and M=cosh d , we have cosh d n+1 R (x) . It follows that R (x) 0 as n for x d (by Equation 10). But d x n n ( n+1 )! was an arbitrary positive number. So by Theorem 8, the series represents cosh x for all x . f. n. 2n. n 2n 2n. 2n. (1) ( x) (1) x x 23. cos x= (1) f (x)=cos ( x)= = n=0 n=0 n=0 (2n)! (2n)! (2n)! . x. 24. e =. n. n. n. , R=. n. ( x/2 ) ( 1 ) x x/2 n , R= f (x)=e = = x n=0 n=0 n n! n! 2 n!. n=0. 2n+1. 2n+1. 2n+2. x 1 n x n x 25. tan x= (1) , R=1 f (x)=xtan x=x (1) = (1) n=0 n=0 n=0 2n+1 2n+1 2n+1 . 1. n. 2n+1. ( 4) 2n+1. n. x 4 n x ( 1 ) 8n+4 26. sin x= (1) , R= f (x)=sin x = (1) = x n=0 n=0 n=0 (2n+1)! (2n+1)! (2n+1)! . x. 27. e =. n. ( ). n. n. n n+2. ( 1 ) x x 2 x 2 ( x ) f (x)=x e =x = n=0 n! n=0 n! n!. n=0. 2n. , R= n 2n. 2n. ( 1 ) 2 x n ( 2x ) 28. cos x= ( 1 ) cos 2x= (1) = n=0 n=0 n=0 (2n)! (2n)! (2n)! . n. f (x)=xcos 2x=. n 2n. ( 1 ) 2. n=0. 2n+1. x. (2n)!. 2n. x . , R=. 29. 1 1 sin x = ( 1cos 2x ) = 2 2 2. =. n=1. . n. 2n. (1) (2x) 1 n=0 (2n)!. 1 = 2. . n. 2n. (1) (2x) 11 n=1 (2n)!. n+1 2n1 2n. (1). 2 x (2n)!. , R=. 30. 1 1 cos x = (1+cos 2x)= 2 2 2. 1+. n=0. n. 2n. ( 1 ) (2x) ( 2n )!. 1 = 2. 1+1+. n=1. n 2n 2n. ( 1 ) 2 x ( 2n )!. 11.

(29) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. =1+. n 2n1 2n. n=1. ( 1 ) 2 x ( 2n )!. , R=. 2. 2. Another method: Use cos x=1sin x and Exercise 29. n 2n+1. n 2n. (1) x sin x 1 (1) x = = 31. and this series also gives the required value at x=0 n=0 (2n+1)! x x n=0 (2n+1)! (namely 1 ); R= .. 32. xsin x x. 3. =. x. n 2n+1. (1) x x n=0 (2n+1)! . 1 3. n 2n+3. =. 1 x. 3. n 2n+1. (1) x xx n=1 (2n+1)! . =. 1 x. 3. n+1 2n+3. (1) x n=0 (2n+3)! . n 2n. (1) x (1) x = = and this series also gives the required value at x=0 (namely 1/6 n=0 (2n+3)! 3 n=0 (2n+3)! x ); R= . . 1. 33. ( n). n. f. 0. ( 1+x ). 1 2 3 4 So f. f. (x) 1/2. ( 0). 1 1 2 1 4 3 8 15 16 . 1 1/2 (1+x) 2 1 3/2 (1+x) 4 3 5/2 (1+x) 8 15 7/2 (1+x) 16 ( n). ( n). n1. (0)=. ( 1 ) 1 3 5 ( 2n3) n. for n 2 , and. 2. 12.

(30) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. n1. n1. (1) 1 3 5 (2n3) n x (1) 1 3 5 (2n3) n 1+x =1+ + x . If a = x , n n n n=2 2 2 n! 2 n! n+1 n a 1 3 5 (2n3)(2n1)x n+1 2 n! then lim =lim n+1 n a n n 2 (n+1)! 1 3 5 (2n3)x n. =. 2n1 x x lim = 2= x <1 for convergence, so R=1 . 2 n n+1 2. Notice that, as n increases, T (x) becomes a better approximation to f (x) for 1<x<1 . n. n. 2. 2n. 2n. 2n. (x ) x x n x n x 34. e = , so e = . Also, cos x= (1) , so = (1) n=0 n! n=0 n! n=0 n=0 n! (2n)! 2 1 1 3 2 13 4 121 6 x n 2n f (x)=e +cos x= (1) + x =2 x + x x + . n=0 n! (2n)! 2 24 720 x. . x. The series for e and cos x converge for all x , so the same is true of the series for f (x) ; that is, R= . From the graphs of f and the first few Taylor polynomials, we see that T (x) provides a closer fit to n. f (x) near 0 as n increases.. 13.

(31) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 2n. n. ( 2) 2n = . ( 1 ) x x 2 35. cos x= (1) f (x)=cos x = n=0 n=0 (2n)! ( 2n )! . n. ( ). n 4n. ( 1 ) x. n=0. (2n)!. , R=. Notice that, as n increases, T (x) becomes a better approximation to f (x) . n. x. ln 2 x. ( ) =e. xln 2. ( xln 2 ). n. . = n=0 n! Notice that, as n increases, T (x) becomes 36. 2 = e. =. . n n. ( ln 2 ) x. n=0. n!. , R= .. n. a better approximation to f (x) .. x. 37. e =. n=0. . 0.2. e. . n. x , so n! n. (0.2) 1 2 1 3 1 4 1 5 1 6 = =10.2+ (0.2) (0.2) + (0.2) (0.2) + (0.2) . But n=0 n! 2! 3! 4! 5! 6! n. 5 (0.2) 1 6 8 0.2 (0.2) =8.8

(32) 10 , so by the Alternating Series Estimation Theorem, e 0.81873 n=0 n! 6! , correct to five decimal places. n 2n+1. ( 1 ) x 38. 3 = radians and sin x= n=0 (2n+1)! 60 . , so 14.

(33) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 60 3!. sin = 60 60. 60 5!. 3. +. 5 3. 5. = + . But 60 1,296,000 93,312,000,000. 5. 8 <10 , so by the Alternating Series Estimation Theorem, 93,312,000,000 3. sin. 0.05234 . 60 60 1,296,000 2n. 3 2n. 6n. x 3 n (x ) n x cos (x )= (1) = (1). 39. cos x= (1) n=0 n=0 n=0 (2n)! (2n)! (2n)! . n. 6n+1. 6n+2. x x 3 n xcos (x )dx=C+ (1) xcos (x )= (1) , with R= . n=0 n=0 (2n)! (6n+2)(2n)! . 3. n. n 2n+1. n 2n. ( 1 ) x sin x 1 ( 1 ) x = = 40. , so n=0 (2n+1)! x x n=0 (2n+1)!. . n 2n. n 2n+1. ( 1 ) x sin x ( 1 ) x dx= dx=C+ n=0 (2n+1)! n=0 (2n+1)(2n+1)! x 3. 41. Using the series from Exercise 33 and substituting x for x , we get. . . 3. x +1 dx =. 3. ( 1 ) x 1+ + n=2 2 4. ( 1 ) x = C+x+ + n=2 8 x. 42. e =. . n=0. n. n. n1. 1 3 5 ( 2n3). x. n. 3n. dx. 2 n!. n1. 1 3 5 ( 2n3) n. x. 3n+1. 2 n!(3n+1) x. n1. x x x e 1 x e 1=. =. n=1 n! n=1 n! n! x. x. n. e 1 x dx=C+ , with R= . n=1 n n! x 6n+2. 1. x 3 43. By Exercise 39, xcos (x )dx=C+ (1) , so xcos (x )dx n=0 (6n+2)(2n)! 0 . 3. 6n+2. 1. n. n. x (1) 1 1 1 1 = = + + , but = (1) n=0 (6n+2)(2n)! 0 n=0 (6n+2)(2n)! 2 8 2! 14 4! 20 6! 1 1 = 0.000069 , so 20 6! 14,400 . n. 15.

(34) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 1. 1. 1. 1. xcos (x )dx 2 16 + 336 0.440 (correct to three decimal places) by the Alternating Series 3. 0. Estimation Theorem. 44. From the table of Maclaurin series in Section .10, we see that 2n+1. 2n+1. x n x tan x= (1) for x in 1,1 and sin x= (1) for all real numbers x , so n=0 n=0 2n+1 (2n+1)! . 1. n. 6n+3. 6n+3. x n x 3 + (1) tan (x )+sin (x )= (1) for x in 1,1 x in 1,1 . Thus, n=0 n=0 2n+1 (2n+1)! 1. 3. 0.2. 0.2. 0. 0. I = dx= . =. . 3. . n. n 6n+3. n=0 (1) x 6n+4. x n=0 (1) 6n+4 . . n. 1 1 + 2n+1 (2n+1)!. 6n+4. (0.2) = (1) n=0 6n+4 n. 10. (0.2) But 10. 1 1 + 3 3!. 1 1 + 2n+1 (2n+1)!. 1 1 + 2n+1 (2n+1)!. dx 0.2 0 4. 10. (0.2) (0.2) = (1+1) 4 10. 1 1 + 3 3!. + .. 10. (0.2) 9 = =5.12

(35) 10 , so by the Alternating Series Estimation Theorem, 20. 4. (0.2) I =0.00080 (correct to five decimal places). 2 45. We first find a series representation for f (x)= ( 1+x ) ( n) ( n) n f (x) f ( 0) 0 1 2 3 . ( 1+x ). 1/2. 1 3/2 (1+x) 2 3 5/2 (1+x) 4 15 7/2 (1+x) 8 . 1/2. , and then substitute.. 1 1 2 3 4 15 8 . 16.

(36) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 2. x 2!. 1 x 3 =1 + 2 4 1+x. 15 8. 3. x 3!. 1. + . 1 3 3 6 5 9 x+ x x + 2 8 16. =1. 1+x. 3. 0.1. . dx 1+x. 0. 1 4 3 7 1 10 = x x + x x + 8 56 32. 3. Estimation Theorem, since. 0.1 0. ( 0.1 ) . 1 4 (0.1) , by the Alternating Series 8. 3 7 8 (0.1) 0.0000000054<10 , which is the maximum desired error. 56. 0.1. Therefore,. . dx 1+x. 0. 0.5. 46.. 2. 2 x. x e 0. 0.5. dx= 0. term with n=2 is. 3. 0.09998750 .. n 2n+2. ( 1 ) x. n=0. n!. dx=. n=0. 1 1 <0.001 , we use n=0 1792. 1/2. n 2n+3. ( 1 ) x. n!(2n+3). ( 1 ). 0. n 2n+3. =. n!(2n+3)2. =. ( 1 ). n=0. n 2n+3. and since the. n!(2n+3)2. 1 1 0.0354 . 24 160. 47. 1. lim x 0. x x. xtan x x. =lim. 3. x 0. =lim x 0. 1 3 1 5 1 7 x + x x + 3 5 7 x. 3. 1 1 2 1 4 x + x 3 5 7. =lim x 0. =. 1 3 1 5 1 7 x x + x 3 5 7 x. 3. 1 3. since power series are continuous functions. 48. lim x 0. 1cos x 1+xe. x. 1 2 1 4 1 6 x+ x x + 2! 4! 6! =lim x 0 1+x 1+x+ 1 x2+ 1 x3+ 1 x4+ 1 x 5+ 1 x6+ 2! 3! 4! 5! 6! 1 1. 17.

(37) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. =lim x 0. . 1 2!. . 1 2!. =lim x 0. 1 2 1 4 1 6 x x + x 2! 4! 6! 2 1 3 1 4 1 5 1 6 x x x x x 3! 4! 5! 6! 1 1 2 1 4 1 x + x 0 2! 4! 2 6! = =1 1 1 2 1 3 1 4 1 x x x x 0 3! 4! 5! 6! 2. since power series are continuous functions. 49. sin xx+ lim x 0. x. 1 3 x 6. 5. x =lim. 1 3 1 5 1 7 x+ x x + 3! 5! 7!. x 0. =lim. x 1 5 1 7 x x + 5! 7!. x 0. x. 5. x+. 1 3 x 6. 5. 2. 4. 1 x x + 5! 7! 9!. =lim x 0. =. 1 1 = 5! 120. since power series are continuous functions. 50. lim x 0. tan xx 3. x+ =lim. 1 3 2 5 x+ x + 3 15. =lim. 3. x 0 x x since power series are continuous functions. 2. 1 3 2 5 x+ x + 3 15. x. 2. x 0. 4. 6. x. 3. =lim x 0. 1 2 2 + x + 3 15. 2. =. 1 3. 4. x x x x x 51. As in Example 8(a), we have e =1 + + and we know that cos x=1 + 1! 2! 3! 2! 4! 2 1 2 1 4 x 2 1 4 1 x + x from Equation 16. Therefore, e cos x= 1x + x . Writing only 2 2 24 the terms with degree 4 , we get 2 1 2 1 4 2 1 4 1 4 3 2 25 4 x e cos x=1 x + x x + x + x + =1 x + x + . 2 24 2 2 2 24 52. x. 18.

(38) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 1+ 1. sec x=. 1 = cos x. 1 2 5 4 x+ x 2 24. 1 2 1 4 x+ x 2 24. 1 1. 1 2 1 4 x+ x 24 2 1 2 1 4 x x + 24 2 1 2 1 4 x x + 4 2 5 4 x + 24 5 4 x + 24 . 1 1 2 5 4 . From the long division above, sec x=1+ x + x + . 1 2 1 4 2 24 x 1 x + 24 2. 53. 1+ x. 1 3 1 5 x+ x 6 120. 1 2 7 4 x+ x + 6 360. x x. 1 3 1 5 x+ x 120 6 1 3 1 5 x x + 120 6 1 3 1 5 x x + 36 6 7 5 x + 360 7 5 x + 360 . 19.

(39) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. x = sin x. x x 1 2 7 4 =1+ x + x + . . From the long division above, 1 3 1 5 sin x 6 360 x x x + 120 6. 54. From Example 6 in Section .9, we have ln (1x)=x x e ln (1x) =. 1 2 1 3 1 2 x + x x x 2 3 2 2 1 3 2 1 3 1 3 x x x x x 3 2 2 2 4 3 x x , x <1 3. 1+x+. 1 2 3 = x 2. = x. ( 4) n =ex. 4n. x x = 55. (1) n=0 n=0 n! n! . 56. . 57. . 58. . n=0. n=0. n=0. 1 2 1 3 x x , x <1 . Therefore, 3 2. n. n 2n. ( 1 ) 2n. 6 (2n)!. =. n 2n+1. ( 1 ) 2n+1. 4. (2n+1)!. n. 3 n. 5 n!. =. . n. n=0. 4 (2n+1)!. n. ( 1 ). n=0. . ( 3/5). n=0. n!. n. =e. , by (11).. 2n 3 6 =cos = , by (16). (2n)! 2 6. (1). =. 4. 3/5. 1. 2n+1. 1 = , by (15). 4 2. =sin. , by (11).. 2. 3. n. 4. n. 3 3 9 27 81 3 3 3 3 3 + + = + + + + = = 1=e 1 , by (11). 59. 3+ + n=1 n! n=0 n! 1! 2! 3! 4! 2! 3! 4!. 60. 1ln 2+. ( ln 2 ). 2. 2!. 61. Assume that f f. / /. (x) f. / /. . ( ln 2 ) 3!. / / /. 3. + =. (x) M , so f. (a) M(xa) f. / /. . ( ln 2 ). n=0 / / /. (x) f. n!. n. ln 2. =e. ( ln 2) 1=21= 12. = e. , by (11). x. (x) M for a x a+d . Now f. / / /. a. / /. x. (a)+M(xa) . Thus, f a. / /. (t)dt . x. (t)dt M dt a. x a. f. / /. (a)+M(ta) dt 20.

(40) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.10 Taylor and Maclaurin Series. 1 1 2 / / / / 2 M(xa) f (x) f (a)+ f (a)(xa)+ M(xa) 2 2 x / x 1 / / / 2 a f (t)dt a f (a)+ f (a)(ta)+ 2 M(ta) dt 1 / / / 2 1 3 f (x) f (a) f (a)(xa)+ f (a)(xa) + M(xa) . So 2 6 1 / / / 2 1 3 f (x) f (a) f (a)(xa) f (a)(xa) M(xa) . But 2 6 1 / / 1 / 2 3 R (x)= f (x)T (x)= f (x) f (a) f (a)(xa) f (a)(xa) , so R (x) M(xa) . 2 2 2 2 6 1 / / / 3 A similar argument using f (x) M shows that R (x) M(xa) . So 2 6 1 3 R x M xa . 2 2 6 Although we have assumed that x>a , a similar calculation shows that this inequality is also true if x<a . /. /. f (x) f (a) f. / /. (a)(xa)+. ( ). 62. (a) f (x)=. {. 2. 1/x. e 0. if x 0 if x=0. so 2. 1/x. /. f (0)=lim x 0. f (x) f (0) e =lim x x0 x 0. =lim x 0. 1/x 2. 1/x. =lim x 0. x 2. =0 (using l’Hospital’s Rule and. 1/x. e 2e simplifying in the penultimate step). Similarly, we can use the definition of the derivative and / / ( 3) ( n) l’Hospital’s Rule to show that f (0)=0 , f (0)=0 , ... , f (0)=0 , so that the Maclaurin series for f consists entirely of zero terms. But since f ( x ) 0 except for x=0 , we see that f cannot equal its Maclaurin series except at x=0 .. (b) From the graph, it seems that the function is extremely flat at the origin. In fact, it could be said to be ‘‘infinitely flat’’ at x=0 , since all of its derivatives are 0 there.. 21.

(41) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 1. The general binomial series in (2) is k(k1) 2 k(k1)(k2) 3 k n x+ x + . ( 1+x ) = n=0 k x =1+kx+ 2! 3! n. ( 1+x ). 1/2. =. 1 2 n. . n=0. 1 2. n. x =1+. 2. 1 2. x+. . 1 2. 2. x+. 2!. 3. 1 2. . 1 2 3!. . 3 2. 3. x + . 4. x 1 3 x 1 3 5 x x = 1+ 2 + 3 + 4 2 2 2! 2 3! 2 4! ( 1 ) x = 1+ + n=2 2. 4. 1. 2.. ( 1+x ). 4. = ( 1+x ) =. . n1. 1 3 5 (2n3)x. n. for x <1,soR=1. n. 2 n!. 4 xn . The binomial coefficient is n. n=0. (4)(5)(6) ( 4n+1 ) (4)(5)(6) (n+3) 4 = = n n! n! n. n. = ( 1 ) 2 3 4 5 6 (n+1)(n+2)(n+3) = (1) (n+1)(n+2)(n+3) 2 3 n!. 1. Thus,. ( 1+x ) 3.. 1. ( 2+x ). 3. 6. n. . (1) (n+1)(n+2)(n+3) n = x for x <1 , so R=1 . n=0 4 6 1. =. 3. =. 1 8. 1+. x 2. 3. x 2. 1 3 8 n=0 n. =. n. . The binomial coefficient is. 2 ( 1+x/2 ) (3)(4)(5) ( 3n+1 ) (3)(4)(5) (n+2) 3 = = n n! n! n. =. n. ( 1 ) 2 3 4 5 (n+1)(n+2) (1) (n+1)(n+2) 2 n! n. =. n. 2 n. (1) (n+1)(n+2)x 1 (1) (n+1)(n+2) x = = Thus, n=0 n=0 3 8 n n+4 2 2 2 ( 2+x ) x <2 , so R=2 .. 1. n. for. x 2. <1. 4. 1.

(42) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 2/3. (1x). =. 2 3 n. n=0. =1+. ( x ). 2 (x)+ 3. 2 3. 1 3. . 2 x2 n=2 3. 2 3. 2. ( x ) +. 2!. ( 1 ) 2 =1 x+ n=2 3. =1. n. n1. . 1 3. . 4 3. 3!. 3. ( x ) + . n. ( 1 ) 2 [1 4 7 ( 3n5) ] n. 3 n! 1 4 7 ( 3n5) n. 3 n!. x. x. n. n. and x <1 x <1 , so R=1 . 5. 4. 18x. = ( 18x ). =1+. =. 1/4. 1 ( 8x ) + 4. =12x+ =12x. . n=0. 1 4. 1 4 n 3 4 2! n. ( 1 ) ( 1 ). ( 8x ). 2. ( 8x ) +. 1 4. n1. . 3 4 3!. n. 3 7 ( 4n5) 8 n. n=2. 4 n!. x. . 7 4. 3. ( 8x ) + . n. n. n=2. and 8x <1 x <. n. 3 7 ( 4n5) 2 n x n!. 1 1 , so R= . 8 8. 6. 1. 5. 1. 1 = 5 = 32x 2 1x/32 2 1 = 2. 1 1+ 5. x 1 32. . x 5. 2. 1/5. n 1 1 x = 5 32 2 n=0 n 1 1 6 6 2 5 5 5 5 x + 10 2! 3! 2. 1 = 2 n=0. +. 1 5 n 11 5. n n. ( 1 ) x 5n. 2. . x. 3. + . 15. 2. 2.

(43) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. =. 1 6 ( 5n4 ) n 1 1 2 1 1 6 1 6 11 3 + x+ x+ x + = + x n 5n+1 n=1 6 2 11 3 16 2 2 52 n! 5 2 5 2! 2 5 3! 2. The radius of convergence is 32 . 7. We must write the binomial in the form ( 1+ expression), so we’ll factor out a 4 . x. x. =. 2. 4+x. (. 2. 4 1+x /4. ). x. =. 2. 1/2. 2. x = 2. x 1+ 4. 2 1+x /4 1 3 2 x 2 x 1 2 = 1+ + 4 2! 2 2 x x n 1 3 5 ( 2n1 ) 2n = + (1) x n n 2 2 n=1 2 4 n!. . 2. 2. x 4. +. 1 2. n. 2. 1 2 n. x = 2 n=0. x 4 3 2 3!. . . 5 2. 3. 2. x 4. + . 2. x x x n 1 3 5 ( 2n1 ) 2n+1 = + (1) x <1 <1 and n=1 3n+1 4 2 2 n! 2. x <2 , so R=2 . 8. 2. 2. 2 x x x = = 2 ( 1+x/2 ) 2 2+x 2. =. x 2 2. 1+ 2. x x = + 2 2 2. x 2. 1 2. . n. n=1(1). 2. 1/2. x 1+ 2. x = 2. +. . n=0. 1 2. . 3 2. 2!. 1 3 5 ( 2n1 ) 2n. x. . 1 2 n. n. x 2. x 2. 2. +. . 1 2. 3 2 3!. . . 5 2. 3. x 2. + . n. n! 2. x n 1 3 5 ( 2n1 ) n+2 = + (1) x and n=1 2n+1/2 2 n! 2. x 2. <1 x <2 , so R=2 .. 9.. 3.

(44) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 3/4. (1+2x). 3 4. . 3 4. 1 4. . 1 4 3!. . 5 4. 3 2 3 (2x)+ (2x) + (2x) + 2! 4 3 n+1 1 5 9 ( 4n7 ) n n =1+ x+3 (1) 2 x n n=2 2 4 n! 3 1 1 n+1 1 5 9 ( 4n7 ) n =1+ x+3 (1) x and 2x <1 x < , so R= . n n=2 2 2 2 2 n! =1+. The three Taylor polynomials are T (x)=1+ 1. 3 3 3 2 3 3 2 5 3 x , T (x)=1+ x x , and T (x)=1+ x x + x 2 3 2 2 8 2 8 16. .. 10. 3. 1/3. 1+4x =(1+4x). 1 3. . 1 3. 2 3. . 2 3 3!. . 5 3. 1 2 3 (4x)+ (4x) + (4x) + 2! 3 4 1 1 n+1 2 5 8 ( 3n4 ) n =1+ x+ (1) (4x) 4x <1 x < R= and , so . n n=2 3 4 4 3 n!. =1+. The three Taylor polynomials are T (x)=1+ 1. T (x)=1+ 3. 4 4 16 2 x , T (x)=1+ x x , and 2 3 3 9. 4 16 2 320 3 x x+ x . 3 9 81. 11. (a) 4.

(45) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 1/2. ( 2). 2. 1/ 1x = 1+ x. 1 2. =1+ =1+. . ( x2) +. 1 2. 3 2. . ( x2) 2+. 2!. . 1 3 5 ( 2n1 ). n=1. 2 n!. n. . 1 2. 3 2 3!. . . 5 2. ( x2) 3+ . 2n. x. (b) 1. sin x. 1. =. 2. . 1 3 5 ( 2n1 ). n=1. (2n+1)2 n!. n. 1x =x+. 12. (a). dx=C+x+. . 1 3 5 ( 2n1 ). n=1. (2n+1)2 n!. n. 2 1/2. ( 1+x ). =. 1. n=0. dx. (b) sinh x=. 2. 2n+1. x. 1. since 0=sin 0=C .. n. sinh x=x+. . =C+x+. . n. 2n+1. ( 1 ) 1 3 5 ( 2n1 ) x 2 n! ( 2n+1 ). n. 1. , but C=0 since sinh 0=0 , so. n. n=1. 2n+1. ( 1 ) 1 3 5 ( 2n1 ) x. , R=1 .. n. n=1. 2n. 1/2 x2n=1+ ( 1 ) 1 3 5 ( 2n1 ) x n n=1 n 2 n!. 1+x 1. 2n+1. x. 2 n! ( 2n+1 ). 13. (a) 3. =(1+x) = 1/3. 1+x. n=0. 1 3 n. 1 3. . . x. n. 1 3. 2 3. . 2 3 3!. 1 2 x+ x+ 2! 3 x n+1 2 5 8 (3n4) n =1+ + (1) x n n=2 3 3 n!. =1+. (b). 3. 1+x =1+. 1 1 2 5 3 x x + x . 3 9 81. 3. . 5 3. 3. x + . 3. 1.01 = 1+0.01 , so let x=0.01 . The sum of the first two. terms is then 5.

(46) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 1 1 2 (0.01) 1.0033 . The third term is (0.01) 0.00001 , which does not affect the fourth decimal 3 9. 1+. place of the sum, so we have. 3. 1.01 1.0033 .. 14. (a) 1/4. 4. 1/ 1+x. =. =(1+x). . . n=0. . 1 4. 1 n 4 x n 5 4 2 x+. . 1 4. 1 x+ 2! 4 1 n 1 5 9 (4n3) n =1 x+ (1) x n n=2 4 4 n!. =1. 5 4 3!. . . 9 4. 3. x + . 1 5 2 15 3 195 4 4 4 x+ x x+ x . 1/ 1.1 =1/ 1+0.1 , so let x=0.1 . The sum of 4 32 128 2048 1 5 2 15 3 the first four terms is then 1 (0.1)+ (0.1) (0.1) 0.976 . The fifth term is 4 32 128 195 4 (0.1) 0.0000095 , which does not affect the third decimal place of the sum, so we have 2048 4. (b) 1/ 1+x =1. 4. 1/ 1.1 0.976 . (Note that the third decimal place of the sum of the first three terms is affected by the fourth term, so we need to use more than three terms for the sum.) 15. (a) 1+ ( x ). 2. =1+(2)(x)+. (2)(3) 2 (2)(3)(4) 3 (x) + (x) + 2! 3!. =1+2x+3x +4x + = 2. so. x. ( 1x ). 2. =x. (b) With x=. . 3. n. (n+1)x =. n=0. . . n. (n+1)x ,. n=0. (n+1)x. n+1. n=0. 1 in part (a), we have n n=1 2. = 1 2. n=1. n. nx . n. =. n=1. n n. 2. =. 1 2 1 1 2. 1 2 = =2 . 2 1 4. 16. (a) 6.

(47) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. 1+ ( x ). ( x+x2). 3. . 3 ( x ) n n=0 n (3)(4) 2 (3)(4)(5) 3 =1+(3)(x)+ (x) + (x) + 2! 3! 3 4 5 ( n+2 ) n 2 3 4 5 ( n+2 ) n =1+ x = x n=1 n=0 n! 2 n! (n+1)(n+2) n = x. n=0 2 =. 1+ ( x ). 3. 3. 3. 2. =x 1+(x) +x 1+(x) (n+1)(n+2) n+1 (n+1)(n+2) n+2 = x + x n=0 n=0 2 2 n(n+1) n n(n+1) n+1 = x + x n=1 n=1 2 2 n(n+1) n (n1)n n n(n+1) (n1)n =x+ x + x =x+ + n=2 n=2 n=2 2 2 2 2 =x+. n=2. 2 n. n x =. n=1. n. 2 n. n x , 1<x<1. 2. (b) Setting x=. x. n 1 in the last series above gives the required series, so = n=1 n 2 2. 1 + 2. 1 2 1 1 2. 2. 3. 3 4 = =6 1 8. . 17. (a) 2 1/2. ( 1+x ). 1 2. =1+. 2. x+. 2. 1 2. ( 1 ) x =1+ + n=2 2. 10. (b) The coefficient of x. . 1 2. 2! n1. ( x2) 2+. 1 3 5 ( 2n3) n. 2 n!. 1 2. . 1 2 3!. . 3 2. ( x2) 3+ . 2n. x. (corresponding to n=5 ) in the above Maclaurin series is. f. ( 10 ). (0) , so 10!. 7.

(48) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. f. ( 10 ). 4. (0) ( 1 ) 1 3 5 7 ( 10 ) =. f (0)=10! 5 10! 2 5!. 1 3 5 7 5. 2 5!. =99 , 225 .. 18. (a) 3 1/2. ( 1+x ). =. . . n=0. =1+ =1+. 1 2 n. 1 2. ( x3) n ( x3) +. . 1 2. . 3 2. (x ) +. 2!. n. ( 1 ) 1 3 5 ( 2n1 ) x. . 3 2. . 1 2. 3 2 3!. . . 5 2. ( x3) 3+ . 3n. n. n=1. 2 n! 9. (b) The coefficient of x (corresponding to n=3 ) in the preceding series is 3 ( 9) ( 9) f (0) f (0) ( 1 ) 1 3 5 9! 5 ( 9) , so =. f (0)= =113 , 400 . 3 9! 9! 8 2 2 3! 19. (a) g(x)=. n=0. k xn g /(x)= k nxn1 , so n=1 n n. k nxn1= k nxn1+ k nxn =(1+x) (1+x)g (x) n=1 n n=1 n n=1 n /. Replace n with n+1 k (n+1)xn+ k nxn n=0 n n=0 n+1 in the first series k(k1)(k2) (kn+1)(kn) n = (n+1) x n=0 ( n+1 )! k(k1)(k2) (kn+1) n + (n) x n=0 n! (n+1)k(k1)(k2) (kn+1) n = kn ) +n x ( n=0 ( n+1 )! k(k1)(k2) (kn+1) n k xn=kg(x) =k x =k n=0 n=0 n! n =. /. Thus, g (x)=. . kg(x) . 1+x. (b) 8.

(49) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.11 The Binomial Series. k. h(x)= ( 1+x ) g(x). /. k1. k. /. g(x)+ ( 1+x ) g (x) [ProductRule k1 k kg(x) =k(1+x) g(x)+ ( 1+x ) 1+x. h (x) =k(1+x). k1. =k(1+x). k1. g(x)+k(1+x). g(x)=0. (c) From part (b) we see that h(x) must be constant for x ( 1,1 ) , so h(x)=h(0)=1 for x ( 1,1 ) . k. k. Thus, h(x)=1= ( 1+x ) g(x)g(x)= ( 1+x ) for x ( 1,1 ) . n1. n. (1) 1 3 5 (2n3)x x 20. By Exercise .11.1, 1+x =1+ + , so n n=2 2 2 n! 1 2 1 3 5 (2n3) 2n 2 1/2 1x =1 x x and n n=2 2 2 n! 1 3 5 (2n3) 2n 1 2 2 2 2 2n 1e sin =1 e sin e sin . Thus, n n=2 2 2 n! /2 /2 1 3 5 (2n3) 2n 1 2 2 2 2 2n 1e sin d =4a 1 e sin e sin L =4a 0 n 0 n=2 2 2 n!. (. ). 2. 1 3 5 (2n3) e S n! 2 2 1 n=2. =4a. /2. where S = n. L = 4a. 0. 2n. sin d =. n. 2. e 2. S. n. 1 3 5 (2n1) by Exercise 44 of 8.1. 2 4 6 2n 2. 2. 2. 1 3 5 (2n3) e 1 1 n=2 2 2 n! 2. 2n. 2. 2n. 2. 2. 2. = 2 a. e e 1 n=2 n 4 4. = 2 a. e 3e 5e 1 4 64 256. 1 3 (2n3) n!. 6. =. n. 1 3 5 (2n1) 2 4 6 2n. 2. = 2 a. 4. 2. e 2. e e 1 3 5 (2n3) (2n1) 1 n n=2 n 4 n! 2 n! 2. 2. d. 2. (2n1). a 2 4 6 (25664e 12e 5e ) 128 9.

(50) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 1. (a) 0 1. ( n) f (x) cos x sin x. 2. cos x. 3. n. f. ( n). T (x). (0). n. 1 0. 1 1. 1. 1. sin x. 0. 1. 4. cos x. 1. 1. 5. sin x. 0. 1. 6. cos x. 1. 1. 1 2 1 2 1 2 1 2 1 2. 2. x. 2. x. 1 4 x 24 2 1 4 x+ x 24 1 6 2 1 4 x+ x x 24 720 2. x+. (b) x 4 2 . T =T. f. 0. T =T. 1. 2. 3. T =T 4. 5. T. 6. 0.7071. 1. 0.6916. 0.7074. 0.7071. 0. 1. 0.2337. 0.0200. 0.0009. 1. 1. 3.9348. 0.1239. 1.2114. (c) As n increases, T (x) is a good approximation to f (x) on a larger and larger interval. n. 2. (a) n. f. ( n). (x). f. ( n). (0). T (x) n. 1.

(51) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 1. 0. x. 1. 1. x. 2. 2x. 3. 6x. 2. 1. 1. 3. 1 ( x1 ) =2x 2. 4. 2. 1 ( x1 ) + ( x1 ) =x 3x+3. 2. 2. 6. 3. 3. 2. 1 ( x1 ) + ( x1 ) ( x1 ) =x +4x 6x+4. (b) x. f. T. 0.9 1.3. 1.1 0.7692. 1 1. 0. T. 1. 1.1 0.7. T. 2. 1.11 0.79. T. 3. 1.111 0.763. (c) As n increases, T (x) is a good approximation to f (x) on a larger and larger interval. n. 3. 0 1. ( n) f (x) ln x 1/x. 2. 1/x. 1. 3. 2/x. 3. 2. 4. 6/x. n. 2. 4. f. ( n). (1). 0 1. 6. 2.

(52) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. T (x)= 4. f. 4 n=0. ( n). (1) 1 n 2 1 3 1 4 (x1) =0+ ( x1 ) (x1) + (x1) (x1) n! 2 3 4. 4. n 0. f e. 1. e. 2. e. 3. e. ( n) x. 3. (2). 2. e. x. 2. e. x. 2. e. x. T (x)=. ( n). f. (x). 2. e. f. 3 n=0. ( n). 2. 2. (2) e n 2 2 2 e 3 (x2) =e +e (x2)+ (x2) + (x2) n! 2 6. 5. n. f. ( n). (x). 0. sin x. 1. cos x. f. ( n). 6. 1 2 3 2 3.

(53) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 2. sin x. . 3. cos x. . T (x)= 3. f. 3. 3 2. 6. ( n). n=0. 1 2. x. n!. 6. n. =. 3 1 + 2 2. x. 6. . 1 4. x. 6. 2. . 3 12. x. 6. 3. 6. ( n). n. f. 0. cos x. . 1. sin x. . 2. cos x. 3. sin x. 4. cos x. T (x)= 4. f. (x). 4 n=0. ( n). 2 3. 1 2 3 2. 1 2 3 2 1 2. f. ( n). 2 3 n!. x. 2 3. n. 4.

(54) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. =. 3 1 2 2. x. 2 3. +. 1 4. x. 2. 2 3. +. 3 12. x. 2 3. 3. . 1 48. x. 2 3. 4. 7. 0. ( n) f (x) arcsinx. 1. 1/ 1x. 2. x/(1x ). 3. (2x +1)/(1x ). n. f 0 2. 1. 2 3/2. 0. 2. T (x)= 3. 3. 2 5/2. f. n=0. ( n). ( n). ( 0). 1. 3. (0) n x x =x+ n! 6. 8. ( n) f (x) (ln x)/x. n 0 1. (1ln x)/x. 2. (3+2ln x)/x. 3. (116ln x)/x. 2. T (x)= 3. f 0. 3 n=0. f. ( n). ( 1). 1 3. 3. 4. 11. ( n). (1) 3 n 2 11 3 (x1) =(x1) (x1) + (x1) n! 2 6. 9. 5.

(55) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. ( n). n. f. 0. xe. 1. (12x)e. 2. 4(x1)e. 3. 4(32x)e. f. (x). 2x. 2x 2x. T (x)= 3. ( 0). 0 2x. f. 3. ( n). n=0. 1 4 12. ( n). (0) n 0 1 1 4 2 12 3 2 3 x = 1+ x + x+ x =x2x +2x n! 1 1 2 6. 10. ( n). n. f. (x). f. 0. ( 3+x2) 1/2 2 1/2 x ( 3+x ) 2 3/2 3 ( 3+x ). 2. 1 2. T (x)= 2. 2 n=0. f. ( n). (1). 1 2 3 8. ( n). (1) 1 3/8 1 3 n 2 2 (x1) =2+ (x1)+ (x1) =2+ (x1)+ (x1) n! 2 2 2 16. 11. In Maple, we can find the Taylor polynomials by the following method: first define f:=sec(x); and then set 6.

(56) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. T2:=convert(taylor(f,x=0,3),polynom);, T4:=convert(taylor(f,x=0,5),polynom);, etc. (The third argument in the taylor function is one more than the degree of the desired polynomial). We must convert to the type polynom because the output of the taylor function contains an error term which we do not want. In Mathematica, we use Tn:=Normal[Series[f,{x,0,n}]] , with n=2,4, etc. Note that in Mathematica, the "degree" argument is the same as the degree of the desired polynomial. In Derive, author sec x , then enter Calculus ,Taylor,8,0; and then simplify the expression. The eighth Taylor polynomial is 1 2 5 4 61 6 277 8 T (x)=1+ x + x+ x+ x . 8 2 24 720 8064. 12. See Exercise 11 for the CAS commands used to generate the Taylor polynomials. The ninth 1 3 2 5 17 7 62 9 Taylor polynomial for tan x is T (x)=x+ x + x+ x+ x . 9 3 15 315 2835. 13. ( n). n. f. 0. x 1 1/2 x 2 1 3/2 x 4 3 5/2 x 8. 1 2 3. (x). ( n) f (4) 2 1 4 1 32. 7.

(57) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. (a) f (x)= x T (x)=2+ 2. (b). R (x) 2. / / /. Since f. R (x) 2. M x4 3!. 3. 1 1/32 1 1 2 2 (x4) (x4) =2+ (x4) (x4) 4 2! 4 64 , where f. / / /. 3. (x) M. Now 4 x 4.2 x4 0.2 x4 0.008 .. (x) is decreasing on 4,4.2 , we can take M= f. 3/256 0.008 (0.008)= =0.000015625 . 6 512. (c) From the graph of R (x) = 2. / / /. (4) =. 3 5/2 3 , so 4 = 8 256 5. x T (x) , it seems that the error is less than 1.52 10 on 4,4.2 2. .. 14. ( n). n. f. 0. x. 1. 2x. 2. 6x. 3. 24x. f. (x). 2. ( n). (1). 1 3. 2. 4. 6 5. (a) 2. f (x) =x T (x) 2. =12(x1)+. 6 2 (x1) 2! 2. =12(x1)+3(x1). (b). R (x) . Since f. 2. / / /. M x1 3!. 3. , where f. / / /. 3. (x) M . Now 0.9 x 1.1 x1 0.1 x1 0.001 .. (x) is decreasing on 0.9,1.1 , we can take 8.

(58) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. / / /. M= f. (0.9) =. 24. ( 0.9 ). , so. 5. 5. R (x) 24/(0.9) (0.001)= 0.004 2 6 0.59049 0.00677404. (c) 2. From the graph of R (x) = x T (x) , it seems that the error is less than 0.0046 on 0.9,1.1 . 2. 2. 15. n 0 1 2 3 4. f. ( n). f. (x). 2/3. ( n). 1 2 3 2 9 8 27. x 2 1/3 x 3 2 4/3 x 9 8 7/3 x 27 56 10/3 x 81 2/3. (a) f (x)=x T (x)=1+ 3. (b). R (x) 3. (1). M x1 4!. 4. 2 2/9 2 1 2 8/27 3 2 4 3 (x1) (x1) + (x1) =1+ (x1) (x1) + (x1) 3 2! 3! 3 9 81 , where. (4). 4. f (x) M . Now 0.8 x 1.2 x1 0.2 x1 0.0016 .. (4). (4). Since f (x) is decreasing on 0.8,1.2 , we can take M= f (0.8) =. 56 10/3 , so (0.8) 81. 56 10/3 (0.8) 81 R (x) (0.0016) 0.00009697 . 3 24 9.

(59) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 2/3. (c) From the graph of R (x) = x T (x) , it seems that the error is less than 0.0000533 on 3. 3. 0.8,1.2 .. 16. ( n). n. f. 0. cos x. 1. sin x. 2. cos x. 3. sin x. 4. cos x. 5. sin x. f. (x). ( n). 3. 1 2 3 2 1 2 3 2 1 2 . (a) f (x)=cos x T 4(x) =. (b). R (x) 4. letting x=. 3 1 2 2 M 5!. x. x 3. 3. . 1 4. x. 5. , where f. 3. ( 5). 1 gives M=1 , so R (x) 4 2 5!. 2. +. 3 12. x. 3. 3. +. (x) M . Now 0 x 3. 1 48. x. 2 3. 3 x. 4. 3. 5. . 3. 5. , and. 5. 0.0105 .. (c). 10.

(60) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. From the graph of R (x) = cos xT (x) , it seems that the error is less than 0.01 on 4. 4. 0,. 2 3. .. 17. n. f. ( n). f 0 1. (x) tan x. 0 1. sec x. 2. 2sec x tan x. 3. 4sec x tan x+2sec x. 4. 8sec x tan x+16sec x tan x. 2. 2 2. 2. 2. 3. 4. 3. since f. ( 4). R (x) 3. M x 4!. 4. 1 3 x 3. , where f. isincreasingon 0, 8. 2. 4. 3. R (x) . 2. 2 3. 6 1 3. ( 4). (x) M . Now 0 x. 4 3 9. 6. 4 x 6. 6. 4. , and letting x=. 6. gives 3. +16 4!. =. (0). 0. (a) f (x)=tan xT (x)=x+. (b). ( n). 2 3. 4. 1 3. 6. 4. 4. 0.057859. (c). From the graph of 11.

(61) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. R (x) = tan xT (3) , it seems that the error is less than 0.006 on 0, /6 . 3. 3. 18. 0. ( n) f (x) ln (1+2x). 1. 2/(1+2x). 2. 4/(1+2x). 3. 16/(1+2x). 4. 96/(1+2x). n. ( n) f (1) ln 3 2 3 4 9 16 27. 2 3 4. (a) f (x)=ln (1+2x)T (x)=ln 3+ 3. 2 4/9 2 16/27 3 (x1) (x1) + (x1) 3 2! 3!. M 4 (4) x1 , where f (x) M . Now 0.5 x 1.5 0.5 x1 0.5 x1 0.5 (b) R (x) 3 4! 1 6 1 1 4 x1 = =0.015625 . , and letting x=0.5 gives M=6 , so R (x) 3 16 4! 16 64. (c) From the graph of R (x) = ln (1+2x)T (x) , it seems that the error is less than 0.005 on 0.5,1.5 . 3. 3. 19. n 0 1 2 3 4. f. ( n). f. (x) 2. e. x. 2. 0. e (2x) 2. 2. 2. e (2+4x ) 2. x. 3. 0. e (12x+8x ) 2. x. (0). 1. x x. ( n). 2. 4. e (12+48x +16x ) 12.

(62) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 2. x. (a) f (x)=e T (x)=1+ 3. R (x) . (b). 3. M x 4!. 4. 2 2 2 x =1+x 2!. , where f. 4. ( 4). (x) M .. 4. Now 0 x 0.1 x ( 0.1 ) , and letting x=0.1 gives 0.01. e. R (x) . ( 12+0.48+0.0016 ). 4. (0.1) 0.00006 .. 24. 3. (c). 2. x. (. 2. From the graph of R (x) = e 1+x 3. ). , it appears that the error is less than 0.000051 on 0,0.1 .. 20. n 0 1 2 3 4. ( n) f (x) x ln x ln x+1 1/x. ( n) f (1) 0 1 1 1. 2. 1/x 2/x. 3. (a) f (x)=xln xT (x)=(x1)+ 3. (b). R (x) 3. M x1 4!. 4. 1 2 1 3 (x1) (x1) 2 6. , where. (4). f (x) M . Now 0.5 x 1.5 x1 . (4). (4). 1 1 4 . x1 2 16 3. f (x) is decreasing on 0.5,1.5 , we can take M= f (0.5) =2/(0.5) =16 , so 16 1 R (x) (1/16)= =0.0416 . 3 24 24 (c) From the graph of R (x) = xln xT (x) , it seems that the error is less than 0.0076 on 0.5,1.5 Since. 3. 3. . 13.

(63) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 21. n 0 1 2 3 4 5. ( n) f (x) xsin x sin x+xcos x 2cos xxsin x 3sin xxcos x 4cos x+xsin x 5sin x+xcos x. ( n) f (0) 0 0 2 0 4. (a) f (x)=xsin xT (x)= 4. (b). R (x) 4. M x 5!. 5. 2 2 4 4 2 1 4 (x0) + (x0) =x x 2! 4! 6. , where. (5). (5). f (x) M . Now 1 x 1 x 1 , and a graph of f (x) shows. 5 5 1 1 = =0.0416 . 4 5! 24 (c) From the graph of R (x) = xsin xT (x) , it seems that the error is less than 0.0082 on 1,1 . that. (5). f (x) 5 for 1 x 1 . Thus, we can take M=5 and get R (x) 4. 4. 22. n 0 1 2 3 4 5. ( n) f (x) sinh 2x 2cosh 2x 4sinh 2x 8cosh 2x 16sinh 2x 32cosh 2x. ( n) f (0) 0 2 0 8 0 32 14.

(64) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 6. 64sinh 2x. (a) f (x)=sinh 2xT (x)=2x+ 5. (b). R (x) 5. M x 6!. 6. 8 3 32 5 4 3 4 5 x+ x =2x+ x + x 3! 5! 3 15. , where. (6). (6). f (x) M . For x in 1,1 , we have x 1 . Since f (x) is an (6). (6). 2. 2. increasing odd function on 1,1 , we see that f (x) f (1)=64sinh 2=32(e e ) 232.119 , so 232.12 6 1 0.3224 . we can take M=232.12 and get R (x) 5 720 (c) From the graph of R (x) = sinh 2xT (x) , it seems that the error is less than 0.027 on 1,1 . 5. 5. 3 1 1 2 3 3 + x x x +R (x) , where 3 2 12 2 6 4 6 6 4 . ( 4) + with f (x) = sin x M=1 . Now x=35 =(30 +5 )= 6 36. 23. From Exercise 5, sin x= M x 3 4! 6 radians, so the error is R (x) . 3 36 3 1. sin 35 + 2 2 R. 4 36 <0.000003 . Therefore, to five decimal places, 4! 2 3 3 1 0.57358 . 12 36 36 36 4. 24. From Exercise 16, 3 1 cos x= x 2 2 3. . 1 4. x. 3. 2. +. 3 12. x. 3. 3. +. 1 48. + radians, the error is R (x) 4 3 20 2 3 3 1 1. + decimal places, cos 69 12 20 20 2 2 4. x=69 =(60 +9 )=. x 20 5! 20. 3. 4. +R (x) . Now since 4. 5 7. <8 10 . Therefore, to five 3. +. 1 48. 20. 4. 0.35837 .. 15.

(65) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. x. e x 25. All derivatives of e are e , so R (x) n (n+1)! x. x. n+1. , where 0<x<0.1 . Letting x=0.1 ,. 0.1. e n+1 (0.1) <0.00001 , and by trial and error we find that n=3 satisfies this inequality R (0.1) n (n+1)! since R (0.1)<0.0000046 . Thus, by adding the four terms of the Maclaurin series for e. x. 3. 0.1. corresponding to n=0 , 1 , 2 , and 3 , we can estimate e. to within 0.00001 . (In fact, this sum is. 0.1. 1.10516 and e 1.10517 .) 26. Example 6 in Section .9 gives the Maclaurin series for ln (1x) as . n. n=1. n. x for x <1 . Thus, n. n. (0.4) n+1 (0.4) = (1) ln 1.4=ln [1(0.4)]= . Since this is an alternating series, the error n=1 n=1 n n is less than the first neglected term by the Alternating Series Estimation Theorem, and we find that 6. a = ( 0.4 ) /6 0.0007<0.001 . So we need the first five (nonzero) terms of the Maclaurin series for 6. the desired accuracy. (In fact, this sum is approximately 0.33698 and ln 1.4 0.33647 .) 1 3 1 5 x+ x . By the Alternating Series Estimation Theorem, the error in the 3! 5! 1 3 1 5 5 1/5 x is less than x <0.01

(66) x <120(0.01)

(67) x < ( 1.2 ) 1.037 . approximation sin x=x 3! 5! 1 3 The curves y=x x and y=sin x0.01 intersect at x 1.043 , so the graph confirms our estimate. 6 Since both the sine function and 27. sin x=x. the given approximation are odd functions, we need to check the estimate only for x>0 . Thus, the desired range of values for x is 1.037<x<1.037 . 28. 16.

(68) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 1 2 1 4 1 6 x+ x x + . By the Alternating Series Estimation Theorem, the error is less 2! 4! 6! 1 6 1 2 1 4 6 1/6 than x <0.005

(69) x <720(0.005)

(70) x < ( 3.6 ) 1.238 . The curves y=1 x + x and 6! 2 24 y=cos x+0.005 intersect at x 1.244 , so the graph confirms our estimate. Since both the cosine function and the given approximation cos x=1. are even functions, we need to check the estimate only for x>0 . Thus, the desired range of values for x is 1.238<x<1.238 . 29. Let s(t) be the position function of the car, and for convenience set s(0)=0 . The velocity of the car is /. / /. v(t)=s (t) and the acceleration is a(t)=s (t) , so the second degree Taylor polynomial is a(0) 2 2 T (t)=s(0)+v(0)t+ t =20t+t . We estimate the distance travelled during the next second to be 2 2 s(1)T (1)=20+1=21 m. The function T (t) would not be accurate over a full minute, since the car 2. 2. 2. could not possibly maintain an acceleration of 2 m / s for that long (if it did, its final speed would be 140 m / s 313 mi / h!) 30. (a) (n). (n). n. (t). 0. e. 1. e. 2. e. (20). ( t20 ). 20. ( t20 ). 20. ( t20 ). 2. 20. . 20. . 20. 2. . 20 /. The linear approximation is T (t)= (20)+ (20)(t20)= 1. 20. 1+ (t20) . The quadratic. approximation is 17.

(71) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. . /. T (t)= (20)+ (20)(t20)+. / /. (20) 1 2 2 2 (t20) = 1+ (t20)+ (t20) 20 2 2. 2. (b). (c). From the graph, it seems that T (t) is within 1% of (t) , that is, 0.99 (t) T (t) 1.01 (t) , for 14 1. 1. C t 58 C. 31. E=. q 2. D. q. . ( D+d ). 2. =. q. q. . 2. 2. D ( 1+d/D ). D. 2. =. q 2. D. We use the Binomial Series to expand ( 1+d/D ) E=. q. 1 12. 2. D =. q. d D. 2. 2. d D. 2 3 + 2! d D. 3. 2. +4. d D d D. 2. .. :. 2. d D. 2 3 4 3!. 3. 2. d D. 1 1+. . . q 2. 3. + . 2. d D. =2qd. D D when D is much larger than d ; that is, when P is far away from the dipole. n. n. 1 32. (a) + = R. 1. o. 2 i. ns. 2 i. i. . 1 3. D. ns. 1 o. (Equation 1) where. o. 18.

(72) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. 2. =. (. ). 2. (. ). 2. R + s +R 2R s +R cos and =. o. o. o. ( ). 2. ( ). R + s R +2R s R cos (2). i. i. i. Using cos 1 gives 2. =. (. ). 2. (. 2. ). R + s +R 2R s +R =. o. o. o. 2. 2. 2. 2. R +s +2Rs +R 2Rs 2R = o. o. s =s o. o. o. and similarly, =s . Thus, Equation 1 becomes i. i. n. n. 1. s. 2. +. s. o. =. 2. (. ). i. 2. (. R + s +R 2R s +R 2. o. 2 i. s. . i. ns. 1 o. s. o. n. 1. s. n n. n. 2. +. s. o. 2. o. ). 1. 2. =. 1. R. i. 2. 1 2 2. 2. 2. 2. 2. R +s +2Rs +R 2Rs +Rs 2R +R = o. . 1 2 in (2) gives us 2. (b) Using cos 1 = o. ns. 1 R. =. o. o. o. 2. 2. 2. 2. s +Rs +R o. o. k. Anticipating that we will use the binomial series expansion ( 1+x ) 1+kx , we can write the last expression for as s o. 1+. o. 2. R R + 2 s s o. 2. and similarly, =s i. 1. i. 2. R R s s2 i. 2. o. n. n. ns. 1 from Equation 1, + = R. 1. o. n. 1. 1+. s. 2. o. n =. 2. R. 1. i. R R + 2 s s o. 2. 2. 2 i. 2. o 2. R R s s2 i. i. 1/2. . i. ns. +. s.

(73) n +n =. 1. 2. . R. 1+. i. Approximating the expressions for . n. 2. R. . s. i. 1/2. 2. R R s s2 i. i. n . 1. R. . s. o.

(74). o. i. n. 1. 2 i. o. i. 1/2. 1. 1 o. n. 2. 1. 1 o. . Thus,. 2. 2. R R + 2 s s o. 1/2. o. 1 o. and . 1. i. by the first two terms in their binomial series, we get 19.

(75) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. n s. o. +. s. i. R R s s2 i. 1 2 1+ 2. 2. R. i. n. 2. 2. R R + 2 s s o. 1 2 1 2. 1. . 2. R R s s2 i. 1 2 1+ 2. 2. o. n =. n. 2. R R + 2 s s o. 1 2 1 2. 1. R. i. n. 1. n 2s. o. o. 2. =. 2. +. s. . 2. R R s s2 i. 2. 2s. i. n. n. 2. 2R. +. i. R R s s2 i. 2. +. R. 2. 2. n. n. o. n. n. o. 2. R R + 2 s s o. 1. . s. 2. 1. R. +. i. 2. 2. R R + 2 s s o. 1. 2R. i. n. o. 2. +. s. s. o. i. n. n. n. 2. 1. . R. R. 2. 1. R. n. 2. R R + 2 s s o. 2s. o. n +. 2. 1. +. n n =. +. 2. 2. R R + 2 s s o. 1. 2. 1. R. =. 1. R. 2R. n. 1 1 + R s. 2s. 2 2. + R. +. o 2. +. 2. R R s s2 i. 2. 2R. . 2. R R s s2 i. 2. n 2. 2s. i. i. 2. o. 1 1 + R s. 1. o. n n 2. R R + 2 s s o. 1. 2. 2. 2. R R s s2 i i. 1 1 R s. i. i. 2 2. n R +. n. 2. 2 2. n n 2. 2. o. o. =.

(76). n. 1. =.

(77). 1 1 + R s. o. n. 1. 2s. o. o. 1 1 + R s. o. 2. +. n. 2. 2s. i. n R +. 1 1 R s. 2. 2s. i. 1 1 R s. i. 1 1 R s. i. 2. i. From Figure 8, we see that sin =h/R . So if we approximate sin with , we get h=R and 2. 2 2. h = R and hence, Equation 4, as desired. 33. (a) If the water is deep, then 2 d/L is large, and we know that tanh x 1 as x . So we can 2. approximate tanh (2 d/L) 1 , and so v gL/(2 )

(78) v gL/(2 ) . (b) From the table, the first term in the Maclaurin series of tanh x is x , so if the water is shallow, we can approximate 20.

(79) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. tanh. 2 d 2 d 2 gL 2 d , and so v

(80) v gd . L L 2 L (n). n. (n). f (0) 0 1. f (x) tanh x. 0 1. 2. sech x 2. 2. 2sech x tanh x. 3. 2sech x(3tanh x1). 2. 0. 2. 2. (c) Since tanh x is an odd function, its Maclaurin series is alternating, so the error in the 2 d 2 d approximation tanh is less than the first neglected term, which is L L f. / / /. 3. 2 d L. (0). 3!. 3. 2 d L. 1 If L>10d , then 3. 3. 2 d L. 1 3. =. 1 < 3. . 3. 2 , so the error in the approximation v =gd is = 375. 3. 1 2 10. 3. gL less than 0.0132gL . 2 375 /2. L g. 34. (a) 4. . dx 2. =4. =4. L g L g. =4. 1k sin x. 0. /2. 2. L g. . 1 3 1 2 2 2 2 1 k sin x + 2! 2. . 1+. 0 /2. (. 1 2. 0. ). 2. (. 1 3 2 4. 2. k sin x+. /2. . (. 2. 2. 1+ k sin x. ). 1/2. dx. 0. 1 3 5 2 2 2 2 2 2 k sin x 3!. ). 4. 4. k sin x+. 1 3 5 2 4 6. ( k2sin 2x) 3+ 6. 6. k sin x+ . dx. dx. L 1 1 3 1 1 3 2 4 + k+ k g 2 2 2 2 4 2 2 2 4 1 3 5 1 3 5 6 + k + 2 4 6 2 2 4 6 [split up the integral and use the result from Exercise 8.1.44] =4. =2. L g. 2. 1+. 1. 2. 2. 2. k+. 2. 2. 2. 2. 13. 2 4. 4. k+. 2. 2. 2. 2. 2. 2. 13 5. 2 4 6. 6. k + 21.

(81) Stewart Calculus ET 5e 0534393217;11. Infinite Sequences and Series; 11.12 Applications of Taylor Polynomials. (b) The first of the two inequalities is true because all of the terms in the series are positive. For the second, T = 2 2. 2. L g. 1+. L g. 1+. 1. 2. 2. 2. k+. 2. 13 2. 2. 2. k+. 2. 2 4. 2. 2. 2. 13 5. 4. 2. 2. 2. k+. 2 4 6. 2. 2. 2. 13 5 7. 6. 2. 2. 2. 2. 2 4 6 8. 8. k + . 1 2 1 4 1 6 1 8 k + k + k + k + 4 4 4 4. The terms in brackets (after the first) form a geometric series with a= . So T 2. L g. 2. 1+. 1 <1 2 0. 2. k /4 1k. 1 2 2 2 k and r=k =sin 4. L 43k . 2 g 44k. =2. 2. (. ). (c) We substitute L=1 , g=9.8 , and k=sin 10 /2 0.08716 , and the inequality from part (b) becomes 2.01090 T 2.01093 , so T 2.0109 . The estimate T 2 L/g 2.0071 differs by about 0.2% .. If =42 , then k 0.35837 and the inequality becomes 2.07153 T 2.08103 , so T 2.0763 . The 0. oneterm estimate is the same, and the discrepancy between the two estimates increases to about 3.4% . 35. (a) L is the length of the arc subtended by the angle , so L=R =L/R . Now sec =(R+C)/R Rsec =R+C C=Rsec R=Rsec (L/R)R .. (b) From Exercise 11, sec xT (x)=1+ 4. L R. 2. L R. 1 2 5 4 x+ x . By part (a), 2 24 2. 4. 2. 4. 4. 1 5 L L L 5L R=R+ R + R R= + . 2 24 4 3 2R 2 R R 24R (c) Taking L=100 km and R=6370 km, the formula in part (a) says that C=Rsec (L/R)R=6370 sec (100/6370)6370 0.78500996544 km. 1 C R 1+ 2. 5 + 24. 2. 4. 2. 4. L 100 5L 5 100 + = + 0.78500995736 km. The formula in part (b) says that C 3 2 6370 3 2R 24R 24 6370 22.