Explanation Given: The function f ( x ) = cos 2 x , a = 0 . Theorem used: Taylor’s remainder theorem The remainder of the Taylor series is R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 for some point c between x and a . Calculation: Obtain the remainder in the Taylor series centered and 0. Substitute 0 for a in R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 , By the theorem stated above, the remainder is R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − 0 ) n + 1 = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x ) n + 1 Therefore, the remainder in the Taylor series is R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x ) n + 1 for some c . To show that lim n → ∞ R n ( x ) = 0 for all x in the interval of convergence. The interval of convergence for Taylor series cos 2 x is ( − ∞ , ∞ ) . Since the ( n + 1 ) th derivative of cos ( 2 x ) is 2 n + 1 times either ± cos 2 x or ± sin 2 x , | f ( n + 1 ) ( c ) | ≤ 2 n + 1

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 9.3, Problem 58E

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To find: The remainder in the in the Taylor series centered at a; show that limn→∞Rn(x)=0 for all x in the interval of convergence.

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Chapter 9.3, Problem 57E

Chapter 9.3, Problem 59E

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Chapter 9 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The remainder in the in the Taylor series centered at a ; show that lim n → ∞ R n ( x ) = 0 for all x in the interval of convergence.

Solution Summary: The author explains that the remainder in the Taylor series centered at a is R_n(x)=

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To find: The remainder in the in the Taylor series centered at a; show that limn→∞Rn(x)=0 for all x in the interval of convergence.

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