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(a) The radius of convergence of ∑ n = 0 ∞ ( n + m m ) ( z ) n by the Cauchy–Hadamard formula. The radius of convergence is R = 1 _ . Formula used: Cauchy Hadamard formula: The radius of convergence of a power series ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . But when series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: The expression ∑ n = 0 ∞ ( n + m m ) ( z ) n is written as: ∑ n = 0 ∞ ( n + m m ) ( z ) n = ∑ n = 0 ∞ ( ( n + m ) ! m ! ( n + m − m ) ! ) ( z ) n = ∑ n = 0 ∞ ( ( n + m ) ! m ! ( n ) ! ) ( z ) n From the given data, it can be observed that a n = ( ( n + m ) ! m ! ( n ) ! ) , a n + 1 = ( ( n + 1 + m ) ! m ! ( n + 1 ) ! ) and z 0 = 0 . Substitute, a n = ( n + m ) ! ( m ) ! ( n ) ! and a n + 1 = ( ( n + 1 + m ) ! m ! ( n + 1 ) ! ) in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! ( n + 1 + m ) ! m ! ( n + 1 ) ! | = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! m ! ( n + 1 ) ! ( n + 1 + m ) ! | = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! m ! ( n + 1 ) ( n ) ! ( n + 1 + m ) ( n + m ) ! | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) | On the further simplification, the following is obtained. R = lim n → ∞ | ( n + 1 ) n + 1 ( 1 + m n + 1 ) | = | 1 ( 1 + m ∞ + 1 ) | = 1 Therefore, the radius of convergence is R = 1 _ . (b) The radius of convergence of ∑ n = 0 ∞ ( n + m m ) ( z ) n from a series of simpler terms. The radius of convergence is R = 1 _ . Result used: Ratio test to find the radius of convergence: If a series with z n ≠ 0 has the property that for every n > N , then | z n + 1 z n | < 1 this series converges absolutely and if | z n + 1 z n | > 1 , then the series diverge. Theorem used: Term wise differentiation of a power series : The derived series of a power series has the same radius of convergence as the original series. Calculation: Differentiating the series expression with respect to z . d d z [ ∑ n = 0 ∞ ( n + m m ) ( z ) n ] = ∑ n = 0 ∞ ( n + m m ) ( n ) ( z ) n − 1 = ∑ n = 0 ∞ ( n + m ) ! m ! n ! ( n ) ( z ) n − 1 = ∑ n = 0 ∞ n ( n + m ) ! m ! n ! ( z ) n − 1 From the given above equation, it can be observed that a n = n ( n + m ) ! m ! n ! and a n + 1 = ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! n ( n + m ) ! m ! n ! | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! m ! n ! n ( n + m ) ! | = lim n → ∞ | ( n + 1 + m ) n | = lim n → ∞ | n ( 1 + 1 + m n ) n | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( 1 + 1 + m n ) | = | ( 1 + 1 + m ∞ ) | = 1 Therefore, the radius of convergence is R = 1 _ .

(a) The radius of convergence of ∑ n = 0 ∞ ( n + m m ) ( z ) n by the Cauchy–Hadamard formula. The radius of convergence is R = 1 _ . Formula used: Cauchy Hadamard formula: The radius of convergence of a power series ∑ n = 0 ∞ a n ( z − z 0 ) n is R = lim n → ∞ | a n a n + 1 | . But when series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: The expression ∑ n = 0 ∞ ( n + m m ) ( z ) n is written as: ∑ n = 0 ∞ ( n + m m ) ( z ) n = ∑ n = 0 ∞ ( ( n + m ) ! m ! ( n + m − m ) ! ) ( z ) n = ∑ n = 0 ∞ ( ( n + m ) ! m ! ( n ) ! ) ( z ) n From the given data, it can be observed that a n = ( ( n + m ) ! m ! ( n ) ! ) , a n + 1 = ( ( n + 1 + m ) ! m ! ( n + 1 ) ! ) and z 0 = 0 . Substitute, a n = ( n + m ) ! ( m ) ! ( n ) ! and a n + 1 = ( ( n + 1 + m ) ! m ! ( n + 1 ) ! ) in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! ( n + 1 + m ) ! m ! ( n + 1 ) ! | = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! m ! ( n + 1 ) ! ( n + 1 + m ) ! | = lim n → ∞ | ( n + m ) ! ( m ) ! ( n ) ! m ! ( n + 1 ) ( n ) ! ( n + 1 + m ) ( n + m ) ! | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) | On the further simplification, the following is obtained. R = lim n → ∞ | ( n + 1 ) n + 1 ( 1 + m n + 1 ) | = | 1 ( 1 + m ∞ + 1 ) | = 1 Therefore, the radius of convergence is R = 1 _ . (b) The radius of convergence of ∑ n = 0 ∞ ( n + m m ) ( z ) n from a series of simpler terms. The radius of convergence is R = 1 _ . Result used: Ratio test to find the radius of convergence: If a series with z n ≠ 0 has the property that for every n > N , then | z n + 1 z n | < 1 this series converges absolutely and if | z n + 1 z n | > 1 , then the series diverge. Theorem used: Term wise differentiation of a power series : The derived series of a power series has the same radius of convergence as the original series. Calculation: Differentiating the series expression with respect to z . d d z [ ∑ n = 0 ∞ ( n + m m ) ( z ) n ] = ∑ n = 0 ∞ ( n + m m ) ( n ) ( z ) n − 1 = ∑ n = 0 ∞ ( n + m ) ! m ! n ! ( n ) ( z ) n − 1 = ∑ n = 0 ∞ n ( n + m ) ! m ! n ! ( z ) n − 1 From the given above equation, it can be observed that a n = n ( n + m ) ! m ! n ! and a n + 1 = ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! n ( n + m ) ! m ! n ! | = lim n → ∞ | ( n + 1 ) ( n + 1 + m ) ! m ! ( n + 1 ) ! m ! n ! n ( n + m ) ! | = lim n → ∞ | ( n + 1 + m ) n | = lim n → ∞ | n ( 1 + 1 + m n ) n | On the further simplification, the following is obtained. lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( 1 + 1 + m n ) | = | ( 1 + 1 + m ∞ ) | = 1 Therefore, the radius of convergence is R = 1 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 15 Solutions

Advanced Engineering Mathematics

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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