Solutions to Taylor and Power Series Problems

Thomas' Calculus: Early Transcendentals

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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 0 . 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 0 . (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 0 (2) Rewrite n tan(1 /n ) as tan(1 /n ) 1 /n . This is a 0 / 0 form and we can use L’Hˆ opital’s Rule to show that the limit is 1. 1
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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. So the quotient will behave like - x 2 / 2 x 2 and go to - 1 / 2 as x goes to 0. (6) Use the Taylor series for cos( x ), substitute x 3 instead of x . Thus we find that cos( x 3 ) - 1 = - x 6 / 2 + higher order terms Similarly, sin( x 2 ) - x 2 = - x 6 3! + higher order terms and so the quotient cos x 3 - 1 sin x 2 - x 2 goes to - x 6 / 2 - x 6 / 6 = 3 as x goes to 0.
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