# On what interval is the expansion valid? Give your answer using interval notation. x 3 1 ? 4 x 6...

## Question:

On what interval is the expansion valid? Give your answer using interval notation.

{eq}\frac{x^3}{1-4x^6} = \sum_{n=0}^{\infty} {/eq}

## Binomial Expansion:

{eq}\\ {/eq}

For a binomial expression with negative indices i.e. Newtonian Binomial Expansion, the restriction is imposed due to convergence of series. Such expansion always has infinite terms provided the exponent of Binomial Expression is a negative but rational number.

The Binomial series expansion for the negative power is 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 {/eq}

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

## Answer and Explanation:

{eq}\\ {/eq}

{eq}\displaystyle f(x) = \dfrac {x^{3}}{1 - 4x^{6}} \\ \displaystyle f(x) = x^{3} \times (1 - 4x^{6})^{-1} {/eq}

The series expansion for the function {eq}\; (1 - 4x^{6})^{-1} \; {/eq} will hold only when {eq}\; |-4x^{6}| < 1 \; {/eq}

So the interval of convergence is given as:

{eq}\; - \; \dfrac {1}{4} < x^{6} < \dfrac {1}{4} \; \; \; \Longrightarrow \boxed{\; \; -\; \biggr( \dfrac {1}{4} \biggr)^{\dfrac {1}{6}} < x < \biggr( \dfrac{1}{4} \biggr)^{\dfrac{1}{6}}} {/eq}