Explanation Given: G ( k ) = ∫ 0 π 2 1 − k 2 sin 2 t . d t Calculation For |k| < 1, |k 2 sin 2 t| < 1 for all t. Substituting −k 2 sin 2 t for t in the binomial series for a = 1/ 2 , we find 1 − k 2 sin 2 t = 1 + ∑ n = 1 ∞ n 1 / 2 ( − k 2 sin 2 t ) n = 1 + ∑ n = 1 ∞ ( − 1 ) n 1 2 ( 1 2 − 1 ) ( 1 2 − 2 ) ... ( 1 2 − n + 1 ) n ! k 2 n sin 2 n t = 1 + ∑ n = 1 ∞ ( − 1 ) n 1. ( 1 − 2 ) . ( 1 − 4 ) ... ( 1 − 2 ( n − 1 ) ) 2 n n ! k 2 n sin 2 n t = 1 + ∑ n = 1 ∞ ( − 1 ) n ( − 1 ) n − 1 1. ( 2 − 1 ) . ( 4 − 1 ) ... ( 2 n − 3 ) 2 n n ! k 2 n sin 2 n t = 1 − ∑ n = 1 ∞ 1.3.5.... ( 2 n − 3 ) 2...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.7, Problem 90E

To determine

To Prove: $G (k) = (\frac{π}{2} - \frac{π}{2} {\sum_{n = 1}^{\infty} (\frac{1.3.... (2 n - 1)}{2.4... (2 n)})}^{2} \frac{{(k)}^{2 n}}{2 n - 1})$

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Chapter 11.7, Problem 89E

Chapter 11.7, Problem 91E

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Chapter 11 Solutions

Calculus - Standalone book

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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To Prove: G ( k ) = ( π 2 − π 2 ∑ n = 1 ∞ ( 1.3.... ( 2 n − 1 ) 2.4... ( 2 n ) ) 2 ( k ) 2 n 2 n − 1 )

Solution Summary: The author explains that k 2 sin 2 t = 1 + n for a = 1/ 2 in the binomial series.

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To Prove: $G (k) = (\frac{π}{2} - \frac{π}{2} {\sum_{n = 1}^{\infty} (\frac{1.3.... (2 n - 1)}{2.4... (2 n)})}^{2} \frac{{(k)}^{2 n}}{2 n - 1})$

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Chapter 11 Solutions

To Prove: G ( k ) = ( π 2 − π 2 ∑ n = 1 ∞ ( 1.3.... ( 2 n − 1 ) 2.4... ( 2 n ) ) 2 ( k ) 2 n 2 n − 1 )

Solution Summary: The author explains that k 2 sin 2 t = 1 + n for a = 1/ 2 in the binomial series.

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To Prove: G(k)=(π2−π2∑n=1∞(1.3....(2n−1)2.4...(2n))2(k)2n2n−1)

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Chapter 11 Solutions

To Prove: $G (k) = (\frac{π}{2} - \frac{π}{2} {\sum_{n = 1}^{\infty} (\frac{1.3.... (2 n - 1)}{2.4... (2 n)})}^{2} \frac{{(k)}^{2 n}}{2 n - 1})$