Explanation Given: The inequality is | cos ( 0.1 ) − T n ( 0.1 ) | ≤ 10 − 7 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. Now, if f ( n + 1 ) exists and is continuous, then the error bound is, | T n ( x ) − f ( x ) | ≤ K | x − a | n + 1 ( n + 1 ) ! ....... (2) Here, K is a number such that | f ( n + 1 ) ( u ) | ≤ K for all u between a and x . Calculation: The value of | f n ( x ) | is either | sin x | or | cos x | depends on n whether n is odd or even then | f ( n ) ( u ) | ≤ 1 for all u . Assume the value of K = 1 and the inequality in equation (2) becomes, | cos ( 0.1 ) − T n ( 0.1 ) | ≤ 1 ⋅ | 0.1 − 0 | n + 1 ( n + 1 ) ! ≤ | 0.1 | n + 1 ( n + 1 ) ! Now, to obtain the value of n to show the above inequality is less than 10 − 7 as, | 0.1 | n + 1 ( n + 1 ) ! ≤ 10 − 7 ....... (3) Now, check for the value of n such that it satisfies the above inequality. Substitute 2 for n in the left hand side of the above inequality (3) as, | 0.1 | 2 + 1 ( 2 + 1 ) ! = | 0.1 | 3 ( 3 ) ! Turn on TI-83 calculator to obtain the value of the above expression and enter the values as shown in Figure 1. Now, press the ENTER to obtain the value of the above expression as shown in Figure 2. For n = 2 , the inequality becomes greater than 10 − 7 . Substitute 3 for n in the left hand side of the above inequality (3) as, | 0.1 | 3 + 1 ( 3 + 1 ) ! = | 0.1 | 4 ( 4 ) ! ≈ 4.17 ⋅ 10 − 6 Turn on TI-83 calculator to obtain the value of the above expression and enter the values as shown in Figure 3. Now, press the ENTER to obtain the value of the above expression as shown in Figure 4. For n = 3 the inequality becomes greater than 10 − 7 . Substitute 4 for n in the left hand side of the above inequality (3) as, | 0.1 | 4 + 1 ( 4 + 1 ) ! = | 0.1 | 5 ( 5 ) ! ≈ 8.33 ⋅ 10 − 8 Turn on TI-83 calculator to obtain the value of the above expression and enter the values as shown in Figure 5. Now, press the ENTER to obtain the value of the above expression as shown in Figure 6. For n = 4 the inequality becomes less than 10 − 7 . Therefore, the value of n such that the inequality | cos ( 0.1 ) − T n ( 0.1 ) | ≤ 10 − 7 is satisfied is 4. Now, verify the above results by calculator. Consider the function f ( x ) = cos x . Differentiate the function f ( x ) = cos x up to degree 4 as, 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 ( − sin x ) = − cos x The third derivative for the function f ( x ) = cos x is calculated as, 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 ( sin x ) = cos x In order to find the Maclaurin polynomial of degree n , use the n th derivative of f ( x ) evaluated at a = 0 as shown in Table 1

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Author: FRANZOSA

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Chapter 11.7, Problem 47E

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To find: The value of n such the inequality |cos(0.1)−Tn(0.1)|≤10−7 is satisfied and the value of a is 0.

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Chapter 11.7, Problem 46E

Chapter 11.7, Problem 48E

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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 value of n such the inequality | cos ( 0.1 ) − T n ( 0.1 ) | ≤ 10 − 7 is satisfied and the value of a is 0.

Solution Summary: The author explains how the nth Taylor polynomial centered at x=a is called Tn. If f(n+1) exists and is continuous, then the error bound is, |Tn(x

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To find: The value of n such the inequality |cos(0.1)−Tn(0.1)|≤10−7 is satisfied and the value of a is 0.

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Chapter 11 Solutions