(a) Explanation Formula used: Taylor polynomial: Let n th degree Taylor polynomial of f at a is denoted by T n ( x ) and is defined as, T n ( x ) = ∑ i = 0 n f ( i ) ( a ) i ! ( x − a ) i = f ( a ) + f ′ ( a ) 1 ! ( x − a ) + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f ( n ) ( a ) n ! ( x − a ) n And f is the sum of its Taylor series, f ( x ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n . Calculation: The given function is f ( x ) = sinh 2 x , centered at a = 0 . Then, f ( 0 ) = 0 . The first derivative of f ( x ) is f ′ ( x ) = 2 cosh 2 x and the corresponding value at a = 0 is f ′ ( 0 ) = 2 . The second derivative is f ″ ( x ) = 4 sinh 2 x and the corresponding value at a = 0 is f ″ ( 0 ) = 0 . The third derivative is f ‴ ( x ) = 8 cosh 2 x and the corresponding value at a = 0 is f ‴ ( 0 ) = 8 . The fourth derivative is f ( 4 ) ( x ) = 16 sinh 2 x and the corresponding value at a = 0 is f ( 4 ) ( 0 ) = 0 . The fifth derivative is f ( 5 ) ( x ) = 32 cosh 2 x and the corresponding value at a = 0 is f ( 5 ) ( 0 ) = 32 (b) To determine To estimate: The accuracy of the approximation f ( x ) ≈ T n ( x ) when x lies in the interval − 1 ≤ x ≤ 1 by using Taylor’s inequality. (c) To determine To check: The result in part (b) by graphing | R n ( x ) | .

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Chapter 11.11, Problem 22E

(a)

To determine

To approximate: The function f(x)=sinh2x by a Taylor polynomial with degree 5 at a=0.

(b)

To determine

To estimate: The accuracy of the approximation f(x)≈Tn(x) when x lies in the interval −1≤x≤1 by using Taylor’s inequality.

(c)

To determine

To check: The result in part (b) by graphing |Rn(x)|.

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Chapter 11.11, Problem 21E

Chapter 11.11, Problem 23E

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Chapter 11 Solutions

CALCULUS: EARLY TRANSCENDENTALS

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To approximate: The function f ( x ) = sinh 2 x by a Taylor polynomial with degree 5 at a = 0 .

Solution Summary: The author calculates the function f(x)=mathrmsinh2x by a Taylor polynomial with degree 5 at a=0.

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To approximate: The function f(x)=sinh2x by a Taylor polynomial with degree 5 at a=0.

To estimate: The accuracy of the approximation f(x)≈Tn(x) when x lies in the interval −1≤x≤1 by using Taylor’s inequality.

To check: The result in part (b) by graphing |Rn(x)|.

Want to see the full answer?

Chapter 11 Solutions