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The region, where the given series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n converges uniformly. The converges uniformly region of series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n is | z | ≤ 7 − δ _ . 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 | . When series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: From the given series function, it can be observed that a n = ( n + 2 7 n − 3 ) n , a n + 1 = ( n + 3 7 n + 4 ) n + 1 and z 0 = i 2 . Substitute, a n = ( n + 2 7 n − 3 ) n and a n + 1 = ( n + 3 7 n + 4 ) n + 1 in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( n + 2 7 n − 3 ) n ( n + 3 7 n + 4 ) n + 1 | = lim n → ∞ | ( n + 2 7 n − 3 ) n × ( 7 n + 4 n + 3 ) n + 1 | = lim n → ∞ | n n 7 n n n ( 1 + 2 n 1 − 3 7 n ) n × 7 n + 1 n n + 1 n n + 1 ( 1 + 4 7 n 1 + 3 n ) n + 1 | = lim n → ∞ | 1 7 n ( 1 + 2 n 1 − 3 7 n ) n × 7 n + 1 1 ( 1 + 4 7 n 1 + 3 n ) n + 1 | On the further simplification, the following is obtained. R = | e 2 e − 3 7 × 7 × e 4 7 e 3 × ( 1 1 ) | = 7 × e 4 + 3 7 − 1 = 7 × e 0 = 7 From the above calculation, it can observed that the series region | z | ≤ 7 − δ , where δ is entirely lies inside the circle of convergence. Therefore, the converges uniformly region of series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n is | z | ≤ 7 − δ _ .

The region, where the given series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n converges uniformly. The converges uniformly region of series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n is | z | ≤ 7 − δ _ . 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 | . When series contain z 2 , the radius of convergence is R = lim n → ∞ | a n a n + 1 | . Calculation: From the given series function, it can be observed that a n = ( n + 2 7 n − 3 ) n , a n + 1 = ( n + 3 7 n + 4 ) n + 1 and z 0 = i 2 . Substitute, a n = ( n + 2 7 n − 3 ) n and a n + 1 = ( n + 3 7 n + 4 ) n + 1 in the equation R = lim n → ∞ | a n a n + 1 | . R = lim n → ∞ | ( n + 2 7 n − 3 ) n ( n + 3 7 n + 4 ) n + 1 | = lim n → ∞ | ( n + 2 7 n − 3 ) n × ( 7 n + 4 n + 3 ) n + 1 | = lim n → ∞ | n n 7 n n n ( 1 + 2 n 1 − 3 7 n ) n × 7 n + 1 n n + 1 n n + 1 ( 1 + 4 7 n 1 + 3 n ) n + 1 | = lim n → ∞ | 1 7 n ( 1 + 2 n 1 − 3 7 n ) n × 7 n + 1 1 ( 1 + 4 7 n 1 + 3 n ) n + 1 | On the further simplification, the following is obtained. R = | e 2 e − 3 7 × 7 × e 4 7 e 3 × ( 1 1 ) | = 7 × e 4 + 3 7 − 1 = 7 × e 0 = 7 From the above calculation, it can observed that the series region | z | ≤ 7 − δ , where δ is entirely lies inside the circle of convergence. Therefore, the converges uniformly region of series f ( z ) = ∑ n = 0 ∞ ( n + 2 7 n − 3 ) n z n is | z | ≤ 7 − δ _ .

ADVANCED ENGINEERING MATH.>CUSTOM<

ADVANCED ENGINEERING MATH.>CUSTOM<

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ISBN: 9781119480150

Author: Kreyszig

Publisher: WILEY C

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ADVANCED ENGINEERING MATH.>CUSTOM<

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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