# 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}

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}