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(a) The radius of convergence of ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n by the Cauchy–Hadamard formula. The radius of convergence is R = 1 5 _ . 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: From the given data, it can be observed that a n = 5 n n ( n + 1 ) and a n + 1 = 5 n + 1 ( n + 1 ) ( n + 2 ) . Substitute, a n = 5 n n ( n + 1 ) and a n + 1 = 5 n + 1 ( n + 1 ) ( n + 2 ) in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | 5 n n ( n + 1 ) 5 n + 1 ( n + 1 ) ( n + 2 ) | = lim n → ∞ | 5 n n ( n + 1 ) ( n + 1 ) ( n + 2 ) 5 n × 5 | = lim n → ∞ | 1 n ( n + 1 ) 5 | = lim n → ∞ | n ( 1 + 1 n ) 5 n | On the further simplification, the following is obtained. R = lim n → ∞ | | ( 1 + 1 n ) 5 | | = ( 1 + 1 ∞ ) 5 = 1 5 Therefore, the radius of convergence is R = 1 5 _ . (b) The radius of convergence of ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n from a series of simpler terms. The radius of convergence is R = 1 5 _ . 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 function ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n with respect to z . d d z [ ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n ] = ∑ n = 1 ∞ 5 n n ( n + 1 ) n ( z ) n − 1 = ∑ n = 1 ∞ 5 n ( n + 1 ) ( z ) n − 1 From the given above equation, it can be observed that a n = 5 n ( n + 1 ) ( z ) n − 1 and a n + 1 = 5 n + 1 ( n + 2 ) ( z ) n . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | 5 n + 1 ( n + 2 ) ( z ) n 5 n ( n + 1 ) ( z ) n − 1 | = lim n → ∞ | 5 n + 1 ( n + 2 ) ( z ) n ( n + 1 ) 5 n ( z ) n − 1 | = lim n → ∞ | 5 n × 5 ( n + 2 ) ( z ) n ( n + 1 ) z 5 n ( z ) n | = lim n → ∞ | 5 z ( n + 1 ) ( n + 2 ) | On the further simplification, the following is obtained. R = lim n → ∞ | 5 z ( 1 + 1 n ) ( 1 + 2 n ) | = | 5 z | By the ratio test the series is convergent if | a n + 1 a n | < 1 . | a n + 1 a n | < 1 | 5 z | < 1 z < 1 5 Therefore, the radius of convergence is R = 1 5 _ .

(a) The radius of convergence of ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n by the Cauchy–Hadamard formula. The radius of convergence is R = 1 5 _ . 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: From the given data, it can be observed that a n = 5 n n ( n + 1 ) and a n + 1 = 5 n + 1 ( n + 1 ) ( n + 2 ) . Substitute, a n = 5 n n ( n + 1 ) and a n + 1 = 5 n + 1 ( n + 1 ) ( n + 2 ) in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | 5 n n ( n + 1 ) 5 n + 1 ( n + 1 ) ( n + 2 ) | = lim n → ∞ | 5 n n ( n + 1 ) ( n + 1 ) ( n + 2 ) 5 n × 5 | = lim n → ∞ | 1 n ( n + 1 ) 5 | = lim n → ∞ | n ( 1 + 1 n ) 5 n | On the further simplification, the following is obtained. R = lim n → ∞ | | ( 1 + 1 n ) 5 | | = ( 1 + 1 ∞ ) 5 = 1 5 Therefore, the radius of convergence is R = 1 5 _ . (b) The radius of convergence of ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n from a series of simpler terms. The radius of convergence is R = 1 5 _ . 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 function ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n with respect to z . d d z [ ∑ n = 1 ∞ 5 n n ( n + 1 ) ( z ) n ] = ∑ n = 1 ∞ 5 n n ( n + 1 ) n ( z ) n − 1 = ∑ n = 1 ∞ 5 n ( n + 1 ) ( z ) n − 1 From the given above equation, it can be observed that a n = 5 n ( n + 1 ) ( z ) n − 1 and a n + 1 = 5 n + 1 ( n + 2 ) ( z ) n . Now the radius of convergence is calculated as: lim n → ∞ | a n + 1 a n | = lim n → ∞ | 5 n + 1 ( n + 2 ) ( z ) n 5 n ( n + 1 ) ( z ) n − 1 | = lim n → ∞ | 5 n + 1 ( n + 2 ) ( z ) n ( n + 1 ) 5 n ( z ) n − 1 | = lim n → ∞ | 5 n × 5 ( n + 2 ) ( z ) n ( n + 1 ) z 5 n ( z ) n | = lim n → ∞ | 5 z ( n + 1 ) ( n + 2 ) | On the further simplification, the following is obtained. R = lim n → ∞ | 5 z ( 1 + 1 n ) ( 1 + 2 n ) | = | 5 z | By the ratio test the series is convergent if | a n + 1 a n | < 1 . | a n + 1 a n | < 1 | 5 z | < 1 z < 1 5 Therefore, the radius of convergence is R = 1 5 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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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

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