# Find the first five terms at the series \delta (x)= \frac{1}{\sqrt [3] {8+x}}

## Question:

Find the first five terms at the series

{eq}\delta (x)= \frac{1}{\sqrt [3] {8+x}}{/eq}

## Binomial Series Representation:

{eq}\\ {/eq}

To get the expression of the power series representation for the given function, we will use the standard Binomial series expansion for the negative fractional powers.

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

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

First of all, we will design our function into a suitable form thus we can apply the Binomial series expansion and get the series representation.

{eq}\\ {/eq}

{eq}\displaystyle f(x) = \dfrac {1}{\sqrt [3] {8 + x}} {/eq}

{eq}\displaystyle f(x) = \dfrac {1}{2} \; \dfrac {1}{\sqrt [3] {1 + \biggr( \dfrac {x}{8} \biggr)}} {/eq}

We know the standard Binomial series representation:

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

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

{eq}\displaystyle (1 + a)^{-\dfrac {1}{3}} = 1 - \dfrac {1}{3} + \dfrac {2a^{2}}{9} - \dfrac {14a^{3}}{81} + \cdots {/eq}

Now substitute the value of {eq}\; a = \dfrac {x}{8} \; {/eq} in the above expression:

{eq}\displaystyle \Biggr[1 + \biggr( \dfrac {x}{8} \biggr) \Biggr]^{-\dfrac {1}{3}} = 1 - \dfrac {1}{3} \; \biggr( \dfrac {x}{8} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {x}{8} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {x}{8} \biggr)^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; |\dfrac {x}{8}| \; < \; 1 \; \; \; \Longrightarrow \; \; |x | < 8 {/eq}

Now finally, the power series representation for the function {eq}\; f(x) \; {/eq} is given as:

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \dfrac {1}{\sqrt [3] {8 + x}} = \dfrac {1}{2} \; \Biggr[ 1 - \dfrac {1}{3} \; \biggr( \dfrac {x}{8} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {x}{8} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {x}{8} \biggr)^{3} + \cdots \Biggr]} {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow |x| < 8 {/eq}