Explanation Given: The function is f ( x ) = e x sin x . Concept used: The n th Taylor polynomial centered at x = a for the function f ( x ) 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: The given function is f ( x ) = e x sin x . Differentiate the function f ( x ) = e x sin x up to degree 3 as, The first derivative for the function f ( x ) = e x sin x is calculated as, f ′ ( x ) = d d x ( e x sin x ) = e x cos x + e x sin x = e x ( cos x + sin x ) The second derivative for the function f ( x ) = e x sin x is calculated as, f ″ ( x ) = d d x ( f ′ ( x ) ) = d d x ( e x ( cos x + sin x ) ) = e x ( cos x + sin x ) + e x ( − sin x + cos x ) = 2 e x cos x The third derivative for the function f ( x ) = e x sin x is calculated as, f ‴ ( x ) = d d x ( f ″ ( x ) ) = d d x ( 2 e x cos x ) = − 2 e x sin x + 2 e x cos x = 2 e x ( cos x sin x ) In order to find the Maclaurin polynomial of degree3, use the first three 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 ) = e x sin x f ( 0 ) = 0 f ′ ( x ) = e x ( cos x + sin x ) f ′ ( 0 ) = 1 f ″ ( x ) = 2 e x cos x f ″ ( 0 ) = 2 f ‴ ( x ) = 2 e x ( cos x sin x ) f ‴ ( x ) = 2 Table 1 Substitute 0 for a and 3 for n in equation (1) to obtain the third Maclaurin polynomial T 3 ( x ) for f ( x ) = e x sin x as, T 3 ( x ) = f ( 0 ) + f ' ( 0 ) 1 ! ( x − 0 ) 1 + f ' ' ( 0 ) 2 ! ( x − 0 ) 2 + f ‴ ( 0 ) 3 ! ( x − 0 ) 3 Substitute 0 for f ( 0 ) , 1 for f ′ ( 0 ) , 2 for f ″ ( 0 ) and 2 for f ‴ ( 0 ) in the above expression as, T 3 ( x ) = 0 + 1 1 ! x + 2 2 ! x 2 + 2 3 ! x 3 = x + 2 2 ⋅ 1 x 2 + 2 3 ⋅ 2 ⋅ 1 x 3 = x + x 2 + 1 3 x 3 The third Maclaurin polynomial for f ( x ) = e x sin x is x + x 2 + 1 3 x 3 . Now, consider the function is f ( x ) = e x . Differentiate the function f ( x ) = e x up to degree 3 as shown below. The first derivative for the function f ( x ) = e x is calculated as, f ′ ( x ) = d d x ( e x ) = e x The second derivative for the function f ( x ) = e x is calculated as, f ″ ( x ) = d d x ( e x ) = e x The third derivative for the function f ( x ) = e x sin x is calculated as, f ‴ ( x ) = d d x ( e x ) = e x In order to find the Maclaurin polynomial of degree 3, use the first three derivative of f ( x ) evaluated at a = 0 as shown in Table 2

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Chapter 11.7, Problem 55E

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To show: The third Maclaurin polynomial for f(x)=exsinx is equal to the product of the third Maclaurin polynomials of f(x)=ex and f(x)=sinx (discarding terms of degree greater than 3 in the product).

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Chapter 11.7, Problem 54E

Chapter 11.7, Problem 56E

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Chapter 11 Solutions

CALCULUS (CLOTH)

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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The third Maclaurin polynomial for f ( x ) = e x sin x is equal to the product of the third Maclaurin polynomials of f ( x ) = e x and f ( x ) = sin x (discarding terms of degree greater than 3 in the product).

Solution Summary: The author explains that the third Maclaurin polynomial for f(x)=exsinx is equal to the product of the second and third versions.

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To show: The third Maclaurin polynomial for f(x)=exsinx is equal to the product of the third Maclaurin polynomials of f(x)=ex and f(x)=sinx (discarding terms of degree greater than 3 in the product).

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Chapter 11 Solutions