Explanation Given: The function is y = cos x , the value of x is π 12 and the value of a is 0. 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: Consider, the given function y = cos x as f ( x ) = cos x . Choose n = 2 and differentiate the function f ( x ) = cos x up to degree 2 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 secondderivative for the function f ( x ) = cos x is calculated as, f ″ ( x ) = d d x ( − sin x ) = − cos x In order to find the Taylor polynomial of degree 2 , use the first two derivative of f ( x ) evaluated at x = 0 as shown below in Table 1. Derivative of function f ( x ) Value of derivative at x = 0 f ( x ) = cos x f ( 0 ) = 1 f ′ ( x ) = − sin x f ′ ( 0 ) = 0 f ″ ( x ) = − cos x f ″ ( 0 ) = − 1 Table 1 The Taylor polynomial of degree 2 is, T 2 ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 Substitute 0 for a in above equation. T 2 ( x ) = f ( 0 ) + f ′ ( 0 ) 1 ! ( x − 0 ) 1 + f ″ ( 0 ) 2 ! ( x − 0 ) 2 Substitute 1 for f ( 0 ) , 0 for f ′ ( 0 ) , and − 1 for f ″ ( 0 ) in above equation

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.4, Problem 26E

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To find: The error |f(x)−T2(x)| at x=π12 for the function y=cosx .

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Chapter 9.4, Problem 25E

Chapter 9.4, Problem 27E

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The error | f ( x ) − T 2 ( x ) | at x = π 12 for the function y = cos x .

Solution Summary: The author explains how to find the Taylor polynomial of degree 2 using the first two derivatives of f(x) evaluated at x=0.

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To find: The error |f(x)−T2(x)| at x=π12 for the function y=cosx .

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Chapter 9 Solutions