Explanation Given information: The differential equation ( 1 − x 2 ) y ″ − x y ′ = 0 . ∑ n = 0 ∞ 2 n ! x 2 n + 1 ( 2 n n ! ) 2 ( 2 n + 1 ) = arcsin x , ( − 1 , 1 ) Proof: Let y = ∑ n = 0 ∞ a n x n be the solution of the differential equation ( 1 − x 2 ) y ″ − x y ′ = 0 . Then, y ′ = ∑ n = 1 ∞ n a n x n − 1 and y ″ = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 . Substituting the value of y and y ″ in the differential equation ( 1 − x 2 ) y ″ − x y ′ = 0 . ( 1 − x 2 ) y ″ − x y ′ = 0 ( 1 − x 2 ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − x ∑ n = 1 ∞ n a n x n − 1 = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − ∑ n = 2 ∞ n ( n − 1 ) a n x n − ∑ n = 1 ∞ n a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 = ∑ n = 2 ∞ n ( n − 1 ) a n x n + ∑ n = 1 ∞

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ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 16.4, Problem 24E

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To prove: That the series ∑n=0∞(2n)!x2n+1(2nn!)2(2n+1) converges to arcsinx on the interval (−1,1) by using the power series solution of the differential equation (1−x2)y″−xy′=0.

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Chapter 16.4, Problem 23E

Chapter 16.4, Problem 25E

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Chapter 16 Solutions

Multivariable Calculus

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To prove: That the series ∑ n = 0 ∞ ( 2 n ) ! x 2 n + 1 ( 2 n n ! ) 2 ( 2 n + 1 ) converges to arcsin x on the interval ( − 1 , 1 ) by using the power series solution of the differential equation ( 1 − x 2 ) y ″ − x y ′ = 0 .

Solution Summary: The author proves that the series displaystyle 'underset'n=0oversetinfty'sum converges to mathrmarcsinx on

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To prove: That the series ∑n=0∞(2n)!x2n+1(2nn!)2(2n+1) converges to arcsinx on the interval (−1,1) by using the power series solution of the differential equation (1−x2)y″−xy′=0.

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Chapter 16 Solutions