Find the Taylor series for f(x) centered at c. f(x) =\frac{1}{1 - x}, \; c = 3 f(x) =...

Question:

Find the Taylor series for f(x) centered at c.

{eq}f(x) =\frac{1}{1 - x}, \; c = 3 {/eq}

{eq}f(x) = \sum_{n=0}^\infty {/eq} _____

Expansion of Binomial Series:

In this case let us assume {eq}k {/eq} is any number and {eq}\left| x \right| < 1 {/eq} then,

{eq}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} k \\ n \\ \end{array}} \right){x^n}} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..... {/eq}

Answer and Explanation:

Here in this case, the given function is:{eq}f(x) =\frac{1}{1 - x}, \; c = 3 {/eq}

Using Binomial expansion, we can rewrite the function as:

{eq}\eqalign{ f(x)& = \frac{1}{{1 - x}} \cr & = \frac{1}{{ - 3 - 1 - x + 3}} \cr & = \frac{1}{{ - 4 - \left( {x - 3} \right)}} \cr & = - \left( {\frac{1}{{1 + \frac{{\left( {x - 3} \right)}}{4}}}} \right) \cr & = - {\left( {1 + \frac{{\left( {x - 3} \right)}}{4}} \right)^{ - 1}} \cr & = - \left( {1 - \frac{{\left( {x - 3} \right)}}{4} + \frac{{{{\left( {x - 3} \right)}^2}}}{{{4^2}}} - \frac{{{{\left( {x - 3} \right)}^3}}}{{{4^3}}} + ....} \right) \cr & = \left( { - 1 + \frac{{\left( {x - 3} \right)}}{4} - \frac{{{{\left( {x - 3} \right)}^2}}}{{{4^2}}} + \frac{{{{\left( {x - 3} \right)}^3}}}{{{4^3}}} - ....} \right) \cr & = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{{\left( {x - 3} \right)}^n}}}{{{4^n}}}} \cr} {/eq}

Which is the require Maclaurin Series expansion of the given function.


Learn more about this topic:

Taylor Series: Definition, Formula & Examples

from AP Calculus AB & BC: Help and Review

Chapter 8 / Lesson 10
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