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The radius of convergence of the given series and identify the sum of the series as a familiar function. The radius of convergence of the given series and power series is R = ∞ _ and ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = sin ( π z ) _ . Series used: A power series in powers of z − z 0 is a series of the form ∑ n = 0 ∞ a n ( z − z 0 ) n = a 0 + a 1 ( z − z 0 ) + a 2 ( z − z 0 ) 2 + ⋅ ⋅ ⋅ . Calculation: The given series is ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! . Compare the series ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! with ∑ n = 0 ∞ a n ( z − z 0 ) n here a n = ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! . Therefore, the value of a n + 1 = ( − 1 ) n + 1 ( π z ) 2 n + 3 ( 2 n + 3 ) ! . According to Cauchy’s-Hadamard formula radius of convergence of ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . Therefore, radius of convergence of series ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! is as follows: R = lim n → ∞ | a n a n + 1 | = lim n → ∞ | ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! ( − 1 ) n + 1 ( π z ) 2 n + 3 ( 2 n + 3 ) ! | = lim n → ∞ | ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! ⋅ ( 2 n + 3 ) ! ( − 1 ) n + 1 ( π z ) 2 n + 3 | = ∞ Hence, the radius of convergence of the given series is R = ∞ . Given series is ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = π z − ( π z ) 2 3 ! + ( π z ) 3 5 ! − ⋅ ⋅ ⋅ = sin ( π z ) called the Maclaurin series Hence, the radius of convergence of the given series and power series is R = ∞ _ and ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = sin ( π z ) _ .

The radius of convergence of the given series and identify the sum of the series as a familiar function. The radius of convergence of the given series and power series is R = ∞ _ and ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = sin ( π z ) _ . Series used: A power series in powers of z − z 0 is a series of the form ∑ n = 0 ∞ a n ( z − z 0 ) n = a 0 + a 1 ( z − z 0 ) + a 2 ( z − z 0 ) 2 + ⋅ ⋅ ⋅ . Calculation: The given series is ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! . Compare the series ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! with ∑ n = 0 ∞ a n ( z − z 0 ) n here a n = ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! . Therefore, the value of a n + 1 = ( − 1 ) n + 1 ( π z ) 2 n + 3 ( 2 n + 3 ) ! . According to Cauchy’s-Hadamard formula radius of convergence of ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . Therefore, radius of convergence of series ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! is as follows: R = lim n → ∞ | a n a n + 1 | = lim n → ∞ | ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! ( − 1 ) n + 1 ( π z ) 2 n + 3 ( 2 n + 3 ) ! | = lim n → ∞ | ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! ⋅ ( 2 n + 3 ) ! ( − 1 ) n + 1 ( π z ) 2 n + 3 | = ∞ Hence, the radius of convergence of the given series is R = ∞ . Given series is ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = π z − ( π z ) 2 3 ! + ( π z ) 3 5 ! − ⋅ ⋅ ⋅ = sin ( π z ) called the Maclaurin series Hence, the radius of convergence of the given series and power series is R = ∞ _ and ∑ n = 0 ∞ ( − 1 ) n ( π z ) 2 n + 1 ( 2 n + 1 ) ! = sin ( π z ) _ .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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ADVANCED ENGINEERING MATH W/ACCESS

Ch. 15.1 - Prob. 1P Ch. 15.1 - Prob. 2P Ch. 15.1 - Prob. 3P Ch. 15.1 - Prob. 4P Ch. 15.1 - Prob. 5P Ch. 15.1 - Prob. 6P Ch. 15.1 - Prob. 7P Ch. 15.1 - Prob. 8P Ch. 15.1 - Prob. 9P Ch. 15.1 - Prob. 10P

Ch. 15.1 - Prob. 12P Ch. 15.1 - Prob. 13P Ch. 15.1 - Prob. 14P Ch. 15.1 - Prob. 15P Ch. 15.1 - Prob. 16P Ch. 15.1 - Prob. 17P Ch. 15.1 - Prob. 18P Ch. 15.1 - Prob. 19P Ch. 15.1 - Prob. 20P Ch. 15.1 - Prob. 21P Ch. 15.1 - Prob. 22P Ch. 15.1 - Prob. 23P Ch. 15.1 - Prob. 24P Ch. 15.1 - Prob. 25P Ch. 15.1 - Prob. 26P Ch. 15.1 - Prob. 27P Ch. 15.1 - Prob. 29P Ch. 15.1 - Prob. 30P Ch. 15.2 - Prob. 1P Ch. 15.2 - Prob. 2P Ch. 15.2 - Prob. 3P Ch. 15.2 - Prob. 4P Ch. 15.2 - Prob. 5P Ch. 15.2 - Prob. 6P Ch. 15.2 - Prob. 7P Ch. 15.2 - Prob. 8P Ch. 15.2 - Prob. 9P Ch. 15.2 - Prob. 10P Ch. 15.2 - Prob. 11P Ch. 15.2 - Prob. 12P Ch. 15.2 - Prob. 13P Ch. 15.2 - Prob. 14P Ch. 15.2 - Prob. 15P Ch. 15.2 - Prob. 16P Ch. 15.2 - Prob. 17P Ch. 15.2 - Prob. 18P Ch. 15.3 - Prob. 1P Ch. 15.3 - Prob. 2P Ch. 15.3 - Prob. 3P Ch. 15.3 - Prob. 4P Ch. 15.3 - Prob. 5P Ch. 15.3 - Prob. 6P Ch. 15.3 - Prob. 7P Ch. 15.3 - Prob. 8P Ch. 15.3 - Prob. 9P Ch. 15.3 - Prob. 10P Ch. 15.3 - Prob. 11P Ch. 15.3 - Prob. 12P Ch. 15.3 - Prob. 13P Ch. 15.3 - Prob. 14P Ch. 15.3 - Prob. 15P Ch. 15.3 - Prob. 16P Ch. 15.3 - Prob. 17P Ch. 15.3 - Prob. 18P Ch. 15.3 - Prob. 19P Ch. 15.4 - Prob. 1P Ch. 15.4 - Prob. 2P Ch. 15.4 - Prob. 3P Ch. 15.4 - Prob. 4P Ch. 15.4 - Prob. 5P Ch. 15.4 - Prob. 6P Ch. 15.4 - Prob. 7P Ch. 15.4 - Prob. 8P Ch. 15.4 - Prob. 9P Ch. 15.4 - Prob. 10P Ch. 15.4 - Prob. 11P Ch. 15.4 - Prob. 12P Ch. 15.4 - Prob. 13P Ch. 15.4 - Prob. 14P Ch. 15.4 - Prob. 16P Ch. 15.4 - Prob. 18P Ch. 15.4 - Prob. 19P Ch. 15.4 - Prob. 20P Ch. 15.4 - Prob. 21P Ch. 15.4 - Prob. 22P Ch. 15.4 - Prob. 23P Ch. 15.4 - Prob. 24P Ch. 15.4 - Prob. 25P Ch. 15.5 - Prob. 2P Ch. 15.5 - Prob. 3P Ch. 15.5 - Prob. 4P Ch. 15.5 - Prob. 5P Ch. 15.5 - Prob. 6P Ch. 15.5 - Prob. 7P Ch. 15.5 - Prob. 8P Ch. 15.5 - Prob. 9P Ch. 15.5 - Prob. 10P Ch. 15.5 - Prob. 11P Ch. 15.5 - Prob. 12P Ch. 15.5 - Prob. 13P Ch. 15.5 - Prob. 14P Ch. 15.5 - Prob. 15P Ch. 15.5 - Prob. 16P Ch. 15.5 - Prob. 17P Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ Ch. 15 - Prob. 3RQ Ch. 15 - Prob. 4RQ Ch. 15 - Prob. 5RQ Ch. 15 - Prob. 6RQ Ch. 15 - Prob. 7RQ Ch. 15 - Prob. 8RQ Ch. 15 - Prob. 9RQ Ch. 15 - Prob. 10RQ Ch. 15 - Prob. 11RQ Ch. 15 - Prob. 12RQ Ch. 15 - Prob. 13RQ Ch. 15 - Prob. 14RQ Ch. 15 - Prob. 15RQ Ch. 15 - Prob. 16RQ Ch. 15 - Prob. 17RQ Ch. 15 - Prob. 18RQ Ch. 15 - Prob. 19RQ Ch. 15 - Prob. 20RQ Ch. 15 - Prob. 21RQ Ch. 15 - Prob. 22RQ Ch. 15 - Prob. 23RQ Ch. 15 - Prob. 24RQ Ch. 15 - Prob. 25RQ Ch. 15 - Prob. 26RQ Ch. 15 - Prob. 27RQ Ch. 15 - Prob. 28RQ Ch. 15 - Prob. 29RQ Ch. 15 - Prob. 30RQ

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