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(a) To show: That ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n by using of Cauchy product. Suppose that f ( z ) = 1 ( 1 − z ) and g ( z ) = 1 ( 1 − z ) from the expression ( 1 − z ) − 2 . f ( z ) = 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = a 0 + a 1 z + a 2 z 2 + a 3 z 3 + ⋅ ⋅ ⋅ From the above calculation, it can be observed that a 0 , a 1 , a 2 , a 3 = 1 . Now the function g ( z ) = 1 ( 1 − z ) is expand as: g ( z ) = 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = b 0 + b 1 z + b 2 z 2 + b 3 z 3 + ⋅ ⋅ ⋅ From the above calculation, it can be observed that b 0 , b 1 , b 2 , b 3 = 1 . Now by use of Cauchy product the value of f ( z ) g ( z ) is calculated as: f ( z ) g ( z ) = ( a 0 + a 1 z + a 2 z 2 + a 3 z 3 + ⋅ ⋅ ⋅ ) ( b 0 + b 1 z + b 2 z 2 + b 3 z 3 + ⋅ ⋅ ⋅ ) 1 ( 1 − z ) 2 = a 0 b 0 + ( a 0 b 1 + a 1 b 0 ) z + ( a 0 b 2 + a 1 b 1 + a 2 b 0 ) z 2 + ⋅ ⋅ ⋅ ( 1 − z ) − 2 = ∑ n = 0 ∞ ( a 0 b n + a 1 b n − 1 + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ a n b 0 ) z n On the further simplification, the following is obtained. ( 1 − z ) − 2 = ∑ n = 0 ∞ ( 1 × 1 + 1 × 1 + ⋅ ⋅ ⋅ + 1 ( n + 1 ) term ) z n ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n Hence proved. (b) To show: That ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n by differentiating a suitable series. The value of 1 ( 1 − z ) is calculated as: 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = ∑ n = 0 ∞ z n Differentiate the function f ( z ) with respect to z . f ′ ( z ) = d d t ( ∑ n = 0 ∞ z n ) − 1 ( 1 − z ) 2 ( − 1 ) = ∑ n = 0 ∞ n z n − 1 1 ( 1 − z ) 2 = ∑ t = 0 ∞ ( t + 1 ) z t ( 1 − z ) − 2 = ∑ t = 0 ∞ ( t + 1 ) z t Hence proved.

(a) To show: That ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n by using of Cauchy product. Suppose that f ( z ) = 1 ( 1 − z ) and g ( z ) = 1 ( 1 − z ) from the expression ( 1 − z ) − 2 . f ( z ) = 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = a 0 + a 1 z + a 2 z 2 + a 3 z 3 + ⋅ ⋅ ⋅ From the above calculation, it can be observed that a 0 , a 1 , a 2 , a 3 = 1 . Now the function g ( z ) = 1 ( 1 − z ) is expand as: g ( z ) = 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = b 0 + b 1 z + b 2 z 2 + b 3 z 3 + ⋅ ⋅ ⋅ From the above calculation, it can be observed that b 0 , b 1 , b 2 , b 3 = 1 . Now by use of Cauchy product the value of f ( z ) g ( z ) is calculated as: f ( z ) g ( z ) = ( a 0 + a 1 z + a 2 z 2 + a 3 z 3 + ⋅ ⋅ ⋅ ) ( b 0 + b 1 z + b 2 z 2 + b 3 z 3 + ⋅ ⋅ ⋅ ) 1 ( 1 − z ) 2 = a 0 b 0 + ( a 0 b 1 + a 1 b 0 ) z + ( a 0 b 2 + a 1 b 1 + a 2 b 0 ) z 2 + ⋅ ⋅ ⋅ ( 1 − z ) − 2 = ∑ n = 0 ∞ ( a 0 b n + a 1 b n − 1 + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ a n b 0 ) z n On the further simplification, the following is obtained. ( 1 − z ) − 2 = ∑ n = 0 ∞ ( 1 × 1 + 1 × 1 + ⋅ ⋅ ⋅ + 1 ( n + 1 ) term ) z n ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n Hence proved. (b) To show: That ( 1 − z ) − 2 = ∑ n = 0 ∞ ( n + 1 ) z n by differentiating a suitable series. The value of 1 ( 1 − z ) is calculated as: 1 ( 1 − z ) = ( 1 − z ) − 1 = 1 + z + z 2 + z 3 + ⋅ ⋅ ⋅ = ∑ n = 0 ∞ z n Differentiate the function f ( z ) with respect to z . f ′ ( z ) = d d t ( ∑ n = 0 ∞ z n ) − 1 ( 1 − z ) 2 ( − 1 ) = ∑ n = 0 ∞ n z n − 1 1 ( 1 − z ) 2 = ∑ t = 0 ∞ ( t + 1 ) z t ( 1 − z ) − 2 = ∑ t = 0 ∞ ( t + 1 ) z t Hence proved.

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

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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