Solutions to Taylor and Power Series Problems

Thomas' Calculus: Early Transcendentals

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Unformatted text preview: Mat104 Solutions to Taylor and Power Series Problems from Old Exams (1) (a). This is a 0 / 0 form. We can use Taylor series to understand the limit. e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + + x n n ! + . . . e- x = 1- x + x 2 2!- x 3 3! + x 4 4! + (- 1) n x n n ! + . . . Thus e x- e- x = 2 x + 2 x 3 3! + 2 x 5 5! + . . . From this we find that e x- e- x- 2 x = 2 x 3 3! + higher degree terms As x approaches 0, the lowest power of x will dominate because the higher degree terms vanish much more rapidly. We can say that e x + e- x- 2 x 2 x 3 3! as x . Next we consider the denominator. x ln(1 + x ) = x ( x- x 2 / 2 + x 3 / 3- x 4 / 4 + . . . ) = x 2- x 3 2 + x 4 3- x 5 4 + . . . . Thus the denominator x 2- x ln(1 + x ) will be dominated by its lowest degree term x 3 2 as we let x 0 and so e x + e- x- 2 x x 2- x ln(1 + x ) 2 x 3 / 3! x 3 / 2 = 4 6 = 2 3 as x . (1b) Again we have a 0 / 0 form. In a similar manner we manipulate Taylor series to determine what power of x the numerator and denominator resemble as x approaches 0. First recall that cos x = 1- x 2 / 2! + x 4 / 4!- x 6 / 6! + . . . and sin x = x- x 3 / 3! + x 5 / 5!- x 7 / 7! + . . . Then we can easily compute that cos x 2- 1 + x 4 / 2 = x 8 / 4! plus higher degree terms x 2 ( x- sin x ) 2 = x 8 / (3!3!) plus higher degree terms Thus cos x 2- 1 + x 4 / 2 x 2 ( x- sin x ) 2 x 8 / 4! x 8 / (3!3!) = 3!3! 4! = 3 2 as x (2) Rewrite n tan(1 /n ) as tan(1 /n ) 1 /n . This is a 0 / 0 form and we can use LH opitals Rule to show that the limit is 1. 1 2 (3) Use the Taylor series for sin x and e x to understand how the numerator behaves near x = 0. (sin x )( e x 2 ) = ( x- x 3 / 3! + x 5 / 5!- . . . )(1 + x 2 + x 4 / 2 + . . . ) = ( x + x 3- x 3 / 3! + higher degree terms ) So sin x e x 2- x = 5 x 3 / 6 + higher degree terms. Now for the denominator. ln(1 + x 3 ) = x 3- ( x 3 ) 2 / 2 + ( x 3 ) 3 / 3- = x 3 + higher degree terms. We conclude that the quotient will go to 5 / 6 as x goes to 0. (4) Here we use the Taylor series for cos x . cos x = 1- x 2 / 2! + x 4 / 4!- x 6 / 6! + . . . = 1- cos x = x 2 / 2!- x 4 / 4! + x 6 / 6!- . . . . So when x is close to 0, 1- cos x x 2 / 2!. When n is large, then 1 /n will be close to 0, so 1- cos(1 /n ) 1 / 2 n 2 . Thus n 2 (1- cos(1 /n )) 1 / 2 as n goes to infinity. (5) Here it is useful to combine the fractions 1 sin x- 1 1- e- x = (1- e- x )- sin x (sin x )(1- e- x ) Again we use power series to understand how the numerator and denominator behave near x = 0. 1- e- x- sin x =- x 2 / 2 + higher order terms (sin x )(1- e- x ) = ( x- x 3 / 3! + x 5 / 5! + . . . )( x- x 2 / 2 + x 3 / 3! + . . . ) = x 2 + higher order terms....
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This homework help was uploaded on 02/12/2008 for the course MATH 104 taught by Professor Nelson during the Fall '07 term at Princeton.

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Solutions to Taylor and Power Series Problems - Mat104...

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