Explanation It is given that 0 < x < 2 c , and we need to prove 1 x = 1 c − 1 c 2 ( x − c ) + 1 c 3 ( x − c ) 2 − 1 c 4 ( x − c ) 3 + ⋯ Recall the common power series expansion of the logarithm function as shown: ln ( 1 + z ) = z − z 2 2 + z 3 3 − z 4 4 + … + ( − 1 ) k + 1 z k k + … ( − 1 < z ≤ 1 ) Plugging z = x c − 1 in the expansion of the logarithm series we get, ln ( 1 + ( x c − 1 ) ) = ( x c − 1 ) − ( x c − 1 ) 2 2 + ( x c − 1 ) 3 3 − ( x c − 1 ) 4 4 + … + ( − 1 ) k + 1 ( x c − 1 ) k k + … ( − 1 < ( x c − 1 ) ≤ 1 ) ⇒ ln ( x c ) = ( x c − 1 ) − ( x c − 1 ) 2 2 + ( x c − 1 ) 3 3 − ( x c − 1 ) 4 4 + … + ( − 1 ) k + 1 ( x c − 1 ) k k + … ( 0 < x ≤ 2 c ) ⇒ ln ( x c ) = ( x c − 1 ) − ( x c − 1 ) 2 2 + ( x c − 1 ) 3 3 − ( x c − 1 ) 4 4 + … + ( − 1 )

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 8, Problem 90SP

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To show: that if 0<x<2c , then 1x=1c−1c2(x−c)+1c3(x−c)2−1c4(x−c)3+⋯

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Chapter 8, Problem 89SP

Chapter 8, Problem 91SP

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Chapter 8 Solutions

Student's Solution and Survival Manual for Calculus

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that if 0 < x < 2 c , then 1 x = 1 c − 1 c 2 ( x − c ) + 1 c 3 ( x − c ) 2 − 1 c 4 ( x − c ) 3 + ⋯

Solution Summary: The author explains that if 0x2c is given, and we need to prove that 1c=fric1,4(x-c)3

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To show: that if 0<x<2c , then 1x=1c−1c2(x−c)+1c3(x−c)2−1c4(x−c)3+⋯

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Chapter 8 Solutions

that if 0 &lt; x &lt; 2 c , then 1 x = 1 c − 1 c 2 ( x − c ) + 1 c 3 ( x − c ) 2 − 1 c 4 ( x − c ) 3 + ⋯

Solution Summary: The author explains that if 0x2c is given, and we need to prove that 1c=fric1,4(x-c)3

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To show: that if 0<x<2c , then 1x=1c−1c2(x−c)+1c3(x−c)2−1c4(x−c)3+⋯

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Chapter 8 Solutions

that if 0 < x < 2 c , then 1 x = 1 c − 1 c 2 ( x − c ) + 1 c 3 ( x − c ) 2 − 1 c 4 ( x − c ) 3 + ⋯