Explanation Given: The period of a pendulum is T = 4 l g ∫ 0 π 2 d θ 1 − k 2 sin 2 θ = 4 l g F ( k ) Where g ≈ 9.8 m/s 2 and k = sin ( θ 0 2 ) . Result used: The Taylor series for ( 1 + x ) p , ( 1 + x ) p = ∑ k = 0 ∞ ( p k ) x k , for | x | < 1 and ( p k ) = p ( p − 1 ) ( p − 2 ) ... ( p − k + 1 ) k ! , ( p 0 ) = 1 The Taylor series for sin x = x − x 3 3 ! + x 5 5 ! − .... Calculation: Taylor series for ( 1 − k 2 sin 2 θ ) − 1 / 2 can be written as ( 1 − k 2 sin 2 θ ) − 1 / 2 = ( 1 − k 2 x 2 ) − 1 / 2 = ( 1 − ( k x ) 2 ) − 1 / 2 The function of ( 1 − ( k x ) 2 ) − 1 / 2 is expressed as binomial series, ( 1 − ( k x ) 2 ) − 1 / 2 = ∑ i = 0 ∞ ( − 1 2 i ) ( k x ) 2 i Substitute x = sin θ in the above expansion, 1 1 − k 2 sin 2 θ = [ 1 + k 2 θ 2 2 + ( − 1 6 k 2 + 3 8 k 4 ) θ 4 + ( 1 45 k 2 − 1 4 k 4 + 5 16 k 6 ) θ 6 + ( − 1 630 k 2 + 3 40 k 4 − 5 16 k 6 + 35 128 k 8 ) θ 8 + ... ] Integrate with respect to θ ∫ 0 π 2 ( 1 1 − k 2 sin 2 θ ) d θ = ∫ ( 1 + k 2 θ 2 2 + ( − 1 6 k 2 + 3 8 k 4 ) θ 4 + ( 1 45 k 2 − 1 4 k 4 + 5 16 k 6 ) θ 6 + ( − 1 630 k 2 + 3 40 k 4 − 5 16 k 6 + 35 128 k 8 ) θ 8 + ... ) d θ = { ∫ 1 d θ + ∫ k 2 θ 2 2 d θ + ∫ ( − 1 6 k 2 + 3 8 k 4 ) θ 4 d θ + ∫ ( 1 45 k 2 − 1 4 k 4 + 5 16 k 6 ) θ 6 d θ + ∫ ( − 1 630 k 2 + 3 40 k 4 − 5 16 k 6 + 35 128 k 8 ) θ 8 d θ + ..

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 11.4, Problem 77E

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To approximate: The value of F(0.1).

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Chapter 11.4, Problem 76E

Chapter 11.4, Problem 78E

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Chapter 11 Solutions

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

Ch. 11.1 - Verify that p3 satisfies p3(k)(a)=f(k)(a), for k =...Ch. 11.1 - Prob. 2QC Ch. 11.1 - Prob. 3QC Ch. 11.1 - Write out the next two Taylor polynomials p4 and...Ch. 11.1 - Prob. 5QC Ch. 11.1 - Prob. 6QC Ch. 11.1 - Suppose you use a second-order Taylor polynomial...Ch. 11.1 - Does the accuracy of an approximation given by a...Ch. 11.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 11.1 - Prob. 4E

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To approximate: The value of F ( 0.1 ) .

Solution Summary: The author calculates the value of F(0.1). The approximation value is 1.574749680

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To approximate: The value of F(0.1).

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Chapter 11 Solutions