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Thomas Calculus 13th [Solutions]

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Section 10.8 Taylor and Maclaurin Series 767<br />

29.<br />

x x x ( n)<br />

x<br />

f ( x) e f ( x) e , f ( x) e f ( x) e ;<br />

2 2 ( n) 2<br />

f (2) e , f (2) e , f (2) e<br />

x<br />

2 3 2<br />

2 2 e 2 e 3<br />

e<br />

2 3! n!<br />

n 0<br />

e e e ( x 2) ( x 2) ( x 2) ( x 2)<br />

n<br />

30.<br />

x x x 2 x 3 ( n)<br />

x n<br />

f ( x) 2 f ( x) 2 ln 2, f ( x) 2 (ln 2) , f ( x) 2 (ln 2) f ( x) 2 (ln 2) ;<br />

2 3 ( n)<br />

f (1) 2, f (1) 2ln 2, f (1) 2(ln 2) , f (1) 2(ln 2) , , f (1) 2(ln 2)<br />

n<br />

x<br />

2 3<br />

2(ln 2) 2 2(ln 2) 3<br />

2(ln 2) ( x 1)<br />

2 3! n!<br />

n 0<br />

2 2 (2ln 2)( x 1) ( x 1) ( x 1)<br />

n<br />

n<br />

31.<br />

32.<br />

f ( x) cos 2 x , f ( x) 2sin 2 x , f ( x) 4cos 2 x , f ( x) 8sin 2 x ,<br />

2 2 2 2<br />

(4) 4 (5) 5<br />

f ( x) 2 cos 2 x , f ( x) 2 sin 2 x , ..;<br />

2 2<br />

(4) 4 (5) (2 n) n 2n<br />

1, 0, 4, 0, 2 , 0, , ( 1) 2<br />

4 4 4 4 4 4 4<br />

f f f f f f f<br />

2 4 n 2n 2n<br />

2<br />

( 1) 2<br />

2 4 3 4<br />

n 0<br />

(2 n)! 4<br />

cos 2 x 1 2 x x x<br />

1 1/2 1 3/2 3 5/2 (4) 15 7/2<br />

2 4 8 16<br />

1 1 3 (4) 15 1 1 2 1 3 5 4<br />

2 4 8 16 2 8 16 128<br />

f ( x) x 1, f ( x) ( x 1) , f ( x) ( x 1) , f ( x) ( x 1) , f ( x) ( x 1) , ;<br />

f (0) 1, f (0) , f (0) , f (0) , f (0) , x 1 1 x x x x<br />

33. The Maclaurin series generated by cos x is<br />

n<br />

n<br />

( 1) 2n<br />

(2 n)!<br />

0<br />

x which converges on ( , ) and the Maclaurin<br />

series generated by<br />

2<br />

2<br />

1 x<br />

2<br />

1 x is n 0<br />

f ( x) cos x is given by<br />

n<br />

x which converges on ( 1, 1). Thus the Maclaurin series generated by<br />

n<br />

( 1) 2n<br />

n<br />

5 2<br />

(2 n)! 2<br />

n 0 n 0<br />

x 2 x 1 2 x x . which converges on the<br />

intersection of ( , ) and ( 1, 1), so the interval of convergence is ( 1, 1).<br />

34. The Maclaurin series generated by<br />

x<br />

e is<br />

n<br />

n<br />

x<br />

n!<br />

0<br />

which converges on ( , ). The Maclaurin series generated<br />

by<br />

2<br />

f ( x) 1 x x e is given by<br />

x<br />

n<br />

2 x 1 2 2 3<br />

n! 2 3<br />

n 0<br />

1 x x 1 x x . which converges on ( , ).<br />

35. The Maclaurin series generated by sin x is<br />

series generated by ln(1 x ) is<br />

( 1)<br />

n<br />

n 1<br />

by f x sin x ln(1 x ) is given by<br />

n 1<br />

( 1) 2n<br />

1<br />

(2n<br />

1)!<br />

n 0<br />

n<br />

n<br />

x which converges on ( , ) and the Maclaurin<br />

x which converges on ( 1, 1). Thus the Maclaurin series generated<br />

n<br />

n 1<br />

( 1) 2n<br />

1 ( 1) n 2 1 3 1 4<br />

(2n<br />

1)! n<br />

2 6<br />

n 0 n 1<br />

x x x x x . which<br />

converges on the intersection of ( , ) and ( 1,1), so the interval convergence is ( 1, 1).<br />

Copyright<br />

2014 Pearson Education, Inc.

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