The Maclaurin Series for the Function f(x) 1/(x-1): A Comprehensive Guide

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The Maclaurin series is a powerful tool in mathematics that allows us to represent a function as an infinite sum of terms involving the derivatives of the function at zero. This guide will explore the Maclaurin series for the function ( f(x) frac{1}{x-1} ), detailing its derivation and applications.

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Deriving the Maclaurin Series Through Binomial Expansion

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Let's start by considering the function ( f(x) frac{1}{1-x} ). We can express this function using a binomial expansion. The binomial theorem for a general power ( alpha ) is given by:

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[ (1 y)^alpha 1 alpha y binom{alpha}{2} y^2 binom{alpha}{3} y^3 cdots binom{alpha}{n} y^n cdots ]

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When ( alpha -1 ) and ( y -x ), we get:

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[ frac{1}{1-x} 1 x x^2 x^3 x^4 cdots ]

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Manipulating the Series to Find ( frac{1}{x-1} )

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To find the series expansion for ( frac{1}{x-1} ), we can manipulate the series for ( frac{1}{1-x} ). By multiplying both sides by -1, we obtain:

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[ frac{1}{x-1} -1 - x - x^2 - x^3 - cdots ]

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Understanding the Geometric Series Representation

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Alternatively, we can use the formula for the sum of an infinite geometric series to represent ( frac{1}{1-x} ). The sum of a geometric series with the first term ( a ) and common ratio ( r ) (where ( |r| " "

[ sum_{n0}^{infty} r^n frac{a}{1-r} ]

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In our case, ( a 1 ) and ( r x ), so:

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[ frac{1}{1-x} sum_{n0}^{infty} x^n ]

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This series converges whenever ( |x| " "

[ frac{1}{x-1} - sum_{n0}^{infty} (-x)^n - sum_{n0}^{infty} x^n ]

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The Derivatives and Maclaurin Series for ( f(x) frac{1}{x-1} )

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To further understand the function, we can calculate its derivatives and apply the Maclaurin series formula. Given:

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[ f(x) frac{1}{1-x} ]

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The first derivative is:

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[ f'(x) frac{1}{(1-x)^2} ]

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The second derivative is:

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[ f''(x) frac{2}{(1-x)^3} ]

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The general pattern for the ( n )-th derivative is:

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[ f^{(n)}(x) frac{n!}{(1-x)^{n 1}} ]

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The Maclaurin series is then given by:

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[ f(x) f(0) frac{f'(0)}{1!}x frac{f''(0)}{2!}x^2 cdots frac{f^{(n)}(0)}{n!}x^n cdots ]

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Since ( f(0) frac{1}{1-0} 1 ) and ( f^{(n)}(0) n! ), the series simplifies to:

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[ f(x) 1 - x - x^2 - x^3 - cdots ]

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Thus, the Maclaurin series for ( f(x) frac{1}{x-1} ) is:

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