Explanation Given: The function is y = e x , the value of x is − 0.5 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 = e x as f ( x ) = e x . Choose n = 2 and differentiate the function f ( x ) = e x up to degree 2 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 secondderivative for the function f ( x ) = e x is calculated as, f ″ ( x ) = d d x ( e x ) = e 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 in Table 1. Derivative of function f ( x ) Value of derivative at x = 0 f ( x ) = e x f ( 0 ) = 1 f ′ ( x ) = e x f ′ ( 0 ) = 1 f ″ ( x ) = e 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

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Author: FRANZOSA

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Chapter 11.7, Problem 31E

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To find: The error |f(x)−T2(x)| at x=−0.5 for the function y=ex .

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Chapter 11.7, Problem 30E

Chapter 11.7, Problem 32E

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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 error | f ( x ) − T 2 ( x ) | at x = − 0.5 for the function y = e 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=−0.5 for the function y=ex .

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Chapter 11 Solutions