Explanation Definition used: The nth Taylor Polynomial: Let f be a function and its first n derivatives are defined at a. Then the n th Taylor polynomial of f at a is the polynomial, P n ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n . Calculation: The function is given by f ( x ) = e − x 2 . Find the sixth Taylor polynomial of the function f ( x ) = e − x 2 at x = 0 as follows by using the above mentioned definition. P n ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n (1) Let f ( x ) = e − x 2 . The derivative f ′ ( x ) is calculated as follows. f ′ ( x ) = − 2 e − x 2 x The derivative f ″ ( x ) is calculated as follows. f ″ ( x ) = − 2 ( − 2 e − x 2 x 2 + e − x 2 ) The derivative f ‴ ( x ) is calculated as follows. f ‴ ( x ) = − 2 d d x ( − 2 e − x 2 x 2 + e − x 2 ) = − 2 ( − 2 ( − 2 e − x 2 x 3 + 2 e − x 2 x ) − 2 e − x 2 x ) = − 2 ( 4 e − x 2 x 3 − 4 e − x 2 x − 2 e − x 2 x ) = − 2 ( 4 e − x 2 x 3 − 6 e − x 2 x ) The derivative f ( 4 ) ( x ) is calculated as follows. f ( 4 ) ( x ) = − 2 d d x ( 4 e − x 2 x 3 − 6 e − x 2 x ) = − 2 ( 4 ( − 2 e − x 2 x 4 + 3 e − x 2 x 2 ) − 6 ( − 2 e − x 2 x 2 + e − x 2 ) ) = − 2 ( − 8 e − x 2 x 4 + 12 e − x 2 x 2 + 12 e − x 2 x 2 − 6 e − x 2 ) = − 2 ( − 8 e − x 2 x 4 + 24 e − x 2 x 2 − 6 e − x 2 ) The derivative f ( 5 ) ( x ) is calculated as follows. f ( 5 ) ( x ) = − 2 d d x ( − 8 e − x 2 x 4 + 24 e − x 2 x 2 − 6 e − x 2 ) = − 2 ( − 8 ( − 2 e − x 2 x 5 + 4 e − x 2 x 3 ) + 24 ( − 2 e − x 2 x 3 + 2 e − x 2 x + 12 e − x 2 x ) ) = − 2 ( 16 e − x 2 x 5 − 32 e − x 2 x 3 − 48 e − x 2 x 3 + 48 e − x 2 x + 12 e − x 2 x ) = − 2 ( 16 e − x 2 x 5 − 80 e − x 2 x 3 + 60 e − x 2 x ) The derivative f ( 6 ) ( x ) is calculated as follows. f ( 6 ) ( x ) = − 2 d d x ( 16 e − x 2 x 5 − 80 e − x 2 x 3 + 60 e − x 2 x ) = − 2 ( 16 ( − 2 e − x 2 x 6 + 5 e − x 2 x 4 ) − 80 ( − 2 e − x 2 x 4 + 3 e − x 2 x 2 ) + 60 ( − 2 e − x 2 x 2 + e − x 2 ) ) = − 2 ( − 32 e − x 2 x 6 + 80 e − x 2 x 4 + 160 e − x 2 x 4 − 240 e − x 2 x 2 − 120 e − x 2 x 2 + 60 e − x 2 ) = − 2 ( 32 e − x 2 x 6 + 240 e − x 2 x 4 − 360 e − x 2 x 2 + 60 e − x 2 ) When x = 0 , find f ( 0 ) as follows

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Chapter 11, Problem 6RE

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To find: The sixth Taylor polynomial of f(x)=e−x2 at x=0 and use it to estimate the value of ∫01e−x2dx accurate to four decimal places.

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Chapter 11, Problem 5RE

Chapter 11, Problem 7RE

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Chapter 11 Solutions

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Ch. 11.1 - Prob. 1CQ Ch. 11.1 - Prob. 2CQ Ch. 11.1 - Prob. 1E Ch. 11.1 - In Exercises 110, find the first three Taylor...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E

Ch. 11.1 - In Exercises 110, find the first three Taylor...Ch. 11.1 - Prob. 10E Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - In Exercises 1120, find the nth Taylor polynomial...Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 17E Ch. 11.1 - Prob. 18E Ch. 11.1 - Prob. 19E Ch. 11.1 - Prob. 20E Ch. 11.1 - Prob. 21E Ch. 11.1 - Prob. 22E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.1 - Prob. 31E Ch. 11.1 - Prob. 32E Ch. 11.1 - Prob. 33E Ch. 11.1 - Prob. 34E Ch. 11.1 - Let f(x) = ex. a. 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The sixth Taylor polynomial of f ( x ) = e − x 2 at x = 0 and use it to estimate the value of ∫ 0 1 e − x 2 d x accurate to four decimal places.

Solution Summary: The author explains how to find the sixth Taylor polynomial of the function f(x)=e-x

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To find: The sixth Taylor polynomial of f(x)=e−x2 at x=0 and use it to estimate the value of ∫01e−x2dx accurate to four decimal places.

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Chapter 11 Solutions