Explanation Given: The function f ( x ) = 1 ( 4 + x 2 ) 2 . Calculation: The first four nonzero terms of the Maclaurin series is computed as follows, f ( x ) = 1 ( 4 + x 2 ) 2 = 1 4 2 ( 1 + x 2 4 ) 2 = 1 16 ( 1 + x 2 4 ) − 2 Substitute x 2 4 for x in ( 1 + x ) − 2 , ( 1 + x 2 4 ) − 2 = 1 − 2 ( x 2 4 ) + 3 ( x 2 4 ) 2 − 4 ( x 2 4 ) 3 + ⋯ = 1 − 2 ( x 2 4 ) + 3 ( x 4 4 2 ) − 4 ( x 6 4 3 ) + ⋯ = 1 − 2 x 2 4 + 3 x 4 4 2

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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ISBN: 9780134996103

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

Publisher: PEARSON

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Chapter 11.3, Problem 59E

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To find: The first four nonzero terms of the Maclaurin series by using the Maclaurin series (1+x)−2=1−2x+3x2−4x3+⋯, for −1<x<1.

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Chapter 11.3, Problem 58E

Chapter 11.3, Problem 60E

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Chapter 11 Solutions

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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The first four nonzero terms of the Maclaurin series by using the Maclaurin series ( 1 + x ) − 2 = 1 − 2 x + 3 x 2 − 4 x 3 + ⋯ , for − 1 < x < 1 .

Solution Summary: The author calculates the first four nonzero terms of the Maclaurin series by using cf(x)=116-

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To find: The first four nonzero terms of the Maclaurin series by using the Maclaurin series (1+x)−2=1−2x+3x2−4x3+⋯, for −1<x<1.

Want to see the full answer?

Chapter 11 Solutions

The first four nonzero terms of the Maclaurin series by using the Maclaurin series ( 1 + x ) − 2 = 1 − 2 x + 3 x 2 − 4 x 3 + ⋯ , for − 1 &lt; x &lt; 1 .

Solution Summary: The author calculates the first four nonzero terms of the Maclaurin series by using cf(x)=116-

Videos

To find: The first four nonzero terms of the Maclaurin series by using the Maclaurin series (1+x)−2=1−2x+3x2−4x3+⋯, for −1<x<1.

Want to see the full answer?

Chapter 11 Solutions

The first four nonzero terms of the Maclaurin series by using the Maclaurin series ( 1 + x ) − 2 = 1 − 2 x + 3 x 2 − 4 x 3 + ⋯ , for − 1 < x < 1 .