Explanation Given: The function is f ( x ) = sin x cos x . Concept used: The n th Taylor polynomial centered at x = a for the function is, T n ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f n ( a ) n ! ( x − a ) n ....... (1) If a = 0 , then T n is also called the n th Maclaurin polynomial. Calculation: First consider the function f ( x ) = sin x . Differentiate the function f ( x ) = sin x up to degree 4 as shown below. The first derivative for the function f ( x ) = sin x is calculated as, f ′ ( x ) = d d x ( sin x ) = cos x The second derivative for the function f ( x ) = sin x is calculated as, f ″ ( x ) = d d x ( f ′ ( x ) ) = d d x ( cos x ) = − sin x The third derivative for the function f ( x ) = sin x is calculated as, f ‴ ( x ) = d d x ( f ″ ( x ) ) = d d x ( − sin x ) = − cos x The fourth derivative for the function f ( x ) = sin x is calculated as, f ( 4 ) ( x ) = d d x ( f ‴ ( x ) ) = d d x ( − cos x ) = sin x In order to obtain the Maclaurin polynomial of degree4, use the first four derivative of f ( x ) evaluated at a = 0 as shown in Table 1. Derivative of function f ( x ) Value of derivative at a = 0 f ( x ) = sin x f ( 0 ) = 0 f ′ ( x ) = cos x f ′ ( 0 ) = 1 f ″ ( x ) = − sin x f ″ ( 0 ) = 0 f ‴ ( x ) = − cos x f ‴ ( x ) = − 1 f ( 4 ) ( x ) = sin x f ( 4 ) ( x ) = 0 Table 1 Substitute 0 for a and 4 for n in equation (1) to obtain the fourth Maclaurin polynomial T 4 ( x ) for f ( x ) = sin x as, T 4 ( x ) = [ f ( 0 ) + f ' ( 0 ) 1 ! ( x − 0 ) 1 + f ' ' ( 0 ) 2 ! ( x − 0 ) 2 + f ‴ ( 0 ) 3 ! ( x − 0 ) 3 + f ( 4 ) ( 0 ) 4 ! ( x − 0 ) 4 ] Substitute 0 for f ( 0 ) , 1 for f ′ ( 0 ) , 0 for f ″ ( 0 ) , − 1 for f ‴ ( 0 ) and 0 for f ( 4 ) ( x ) in the above expression as, T 4 ( x ) = 0 + 1 1 ! x + 0 2 ! x 2 + − 1 3 ! x 3 + 0 4 ! ( x − 0 ) 4 = x + 0 ⋅ x 2 + − 1 3 ⋅ 2 ⋅ 1 x 3 + 0 ⋅ x 4 = x − 1 6 x 3 The fourth Maclaurin polynomial for f ( x ) = sin x is x − 1 6 x 3 . Now consider the function f ( x ) = cos x . Differentiate the function f ( x ) = cos x up to degree 4 as shown below. The first derivative for the function f ( x ) = cos x is calculated as, f ′ ( x ) = d d x ( cos x ) = − sin x The second derivative for the function f ( x ) = cos x is calculated as, f ″ ( x ) = d d x ( f ′ ( x ) ) = d d x ( − sin x ) = − cos x The third derivative for the function f ( x ) = cos x is calculated as, f ‴ ( x ) = d d x ( f ″ ( x ) ) = d d x ( − cos x ) = sin x The fourth derivative for the function f ( x ) = cos x is calculated as, f ( 4 ) ( x ) = d d x ( f ‴ ( x ) ) = d d x ( sin x ) = cos x In order to obtain the Maclaurin polynomial of degree4, use the first four derivative of f ( x ) evaluated at a = 0 as shown in Table 2

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.4, Problem 50E

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To find: The fourth Maclaurin polynomial for f(x)=sinxcosx by multiplying the fourth Maclaurin polynomials for f(x)=sinx and f(x)=cosx .

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Chapter 9.4, Problem 49E

Chapter 9.4, Problem 51E

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Chapter 9 Solutions

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Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 4PQ Ch. 9.1 - Prob. 5PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E

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The fourth Maclaurin polynomial for f ( x ) = sin x cos x by multiplying the fourth Maclaurin polynomials for f ( x ) = sin x and f ( x ) = cos x .

Solution Summary: The author explains how to obtain the fourth Maclaurin polynomial for f(x)=sinxcosx by multiplying the nth Taylor.

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To find: The fourth Maclaurin polynomial for f(x)=sinxcosx by multiplying the fourth Maclaurin polynomials for f(x)=sinx and f(x)=cosx .

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Chapter 9 Solutions