Explanation Formula used: 1. Maclaurin expansion of ln ( 1 + x ) : “The Maclaurin expansion of ln ( 1 + x ) is given by, ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ⋯ ” 2. Taylor series expansion of a function: f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) ( x − a ) 2 2 ! + … . Calculation: Replace x by x − 1 in the Maclaurin expansion of ln ( 1 + x ) and obtain the Maclaurin expansion of ln x as shown below ln x = ( x − 1 ) − ( x − 1 ) 2 2 + ( x − 1 ) 3 3 − ( x − 1 ) 4 4 + ⋯ Now consider the function f ( x ) = ln x with a = 1 . The derivatives and its values are obtained as follows. Differentiate the function f ( x ) = ln x with respect to x as shown below. f ′ ( x ) = d d x ( ln x ) = 1 x Again, differentiate the function f ′ ( x ) with respect to x as follows. f ″ ( x ) = d d x ( 1 x ) = − 1 ( x − 2 ) = − 1 x 2 Now differentiate the function f ″ ( x ) with respect to x as shown below. f ‴ ( x ) = d d x ( − 1 x 2 ) = − 1 ( − 2 ) x − 3 = 2 x − 3 = 2 x 3 Similarly obtain f i v ( x ) as shown below. f i v ( x ) = d d x ( 2 x 3 ) = 2 ( − 3 ) x − 4 = − 6 x − 4 = − 6 x 4 Substitute x = 1 in f ( x ) and obtain f ( 1 ) as shown below

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Chapter 30, Problem 61RE

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The power series for lnx by replacing x with (x−1) in the power series of ln(1+x) and compare it with the Taylor series expansion of lnx with a=1.

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Chapter 30, Problem 60RE

Chapter 30, Problem 62RE

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Chapter 30 Solutions

Basic Technical Mathematics

Ch. 30.1 - Prob. 1PE Ch. 30.1 - Prob. 2PE Ch. 30.1 - Prob. 1E Ch. 30.1 - Prob. 2E Ch. 30.1 - Prob. 3E Ch. 30.1 - Prob. 4E Ch. 30.1 - Prob. 5E Ch. 30.1 - Prob. 6E Ch. 30.1 - Prob. 7E Ch. 30.1 - Prob. 8E

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The power series for ln x by replacing x with ( x − 1 ) in the power series of ln ( 1 + x ) and compare it with the Taylor series expansion of ln x with a = 1 .

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The power series for lnx by replacing x with (x−1) in the power series of ln(1+x) and compare it with the Taylor series expansion of lnx with a=1.

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Chapter 30 Solutions