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Find the Maclaurin series for the function f (x) = {1} / {square root {4 + x^2}}. What its radius...

Question:

Find the Maclaurin series for the function {eq}\displaystyle f (x) = \frac {1} {\sqrt {4 + x^2}} {/eq}. What its radius of convergence?

Maclaurin Series Representation:

To obtain the power series representation and the radius of convergence for the given function, we will use the Binomial series expansion for negative fractional powers. Firstly we will convert the given function into the standard form of Binomial series expression then we will expand it.

{eq}(1 + a)^{n} = 1 + na + \dfrac {n (n - 1)}{2!} \; a^{2} + \dfrac {n (n - 1) (n - 2)}{3!} \; a^{3} + \cdots {/eq}

The above series representation holds only when: {eq}\; \; |a| < 1 {/eq}

Answer and Explanation:

Notice that

{eq}f(x) = \dfrac {1}{\sqrt {4 + x^{2}}} {/eq}

{eq}\Rightarrow f(x) = \dfrac {\dfrac {1}{2}}{\sqrt {1 + \biggr( \dfrac {x}{2} \biggr)^{2}}} = \biggr( \dfrac {1}{2} \biggr) \; \Biggr[1 + \biggr( \dfrac {x}{2} \biggr)^{2} \Biggr]^{-\dfrac {1}{2}} {/eq}


We know the standard Binomial series representation for the fractional negative power given as:

{eq}(1 + a)^{n} = 1 + na + \dfrac {n(n - 1)}{2!} \; a^{2} + \dfrac {n (n - 1)(n - 2)}{3!} \; a^{3} + \cdots \quad \quad \quad (1) {/eq}


The power series in equation (1) holds when: {eq}\; \; |a| < 1 {/eq}

{eq}(1 + a)^{-\dfrac {1}{2}} = 1 - \dfrac {a}{2} + \dfrac {\biggr( - \dfrac {1}{2} \biggr) \; \biggr( - \dfrac {1}{2} - 1 \biggr)}{2!} \; a^{2} + \dfrac {\biggr( - \dfrac {1}{2} \biggr) \; \biggr( - \dfrac {1}{2} - 1 \biggr) \; \biggr( - \dfrac {1}{2} - 2 \biggr)}{3!} \; a^{3} + \cdots {/eq}

{eq}\Rightarrow (1 + a)^{-\dfrac {1}{2}} = 1 - \dfrac {a}{2} + \dfrac {3a^{2}}{8} - \dfrac {5 a^{3}}{16} + \cdots {/eq}


Now replace the value of {eq}\; a = \biggr( \dfrac {x}{2} \biggr)^{2} \; {/eq} in the above expression:

{eq}\Biggr[1 + \biggr( \dfrac {x}{2} \biggr)^{2} \Biggr]^{-\dfrac {1}{2}} = 1 - \dfrac {\biggr( \dfrac {x}{2} \biggr)^{2}}{2} + \dfrac {3}{8} \; \biggr( \dfrac {x}{2} \biggr)^{4} - \dfrac {5}{16} \; \biggr( \dfrac {x}{2} \biggr)^{6} + \cdots {/eq}

The above series representation holds only when: {eq}\; |\biggr( \dfrac {x}{2} \biggr)^{2}| < 1 \; \; \Rightarrow |x| < 2 {/eq}


Now finally the power series representation for the main function is given as:

{eq}\Rightarrow \boxed {\dfrac {1}{\sqrt {4 + x^{2}}} = \dfrac {1}{2} \; \Biggr[ 1 - \dfrac {1}{2} \; \biggr( \dfrac {x}{2} \biggr)^{2} + \dfrac {3}{8} \; \biggr( \dfrac {x}{2} \biggr)^{4} - \dfrac {5}{16} \; \biggr( \dfrac {x}{2} \biggr)^{6} + \cdots \Biggr]} {/eq}

The radius of convergence is given as: {eq}\; \; R = 2 {/eq}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
9.8K

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