# Use the binomial series to expand the function as a power series. 5/(4 + x)^3 State the radius of...

## Question:

Use the binomial series to expand the function as a power series:

{eq}\displaystyle \frac{5}{(4 + x)^{3}} {/eq}

State the radius of convergence R.

## Binomial Series Expansion:

{eq}\\ {/eq}

To expand the given function as a power series representation, we will use the binomial series expansion for the fractional negative powers. First of all, we will determine the power series for the standard binomial function then we will go for the natural function with the help of substitution.

{eq}\displaystyle (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 good only when: {eq}\; \; \Longrightarrow \; \; |a| < 1 {/eq}

## Answer and Explanation:

{eq}\\ {/eq}

{eq}f(x) = \dfrac {5}{(4 + x)^{3}} {/eq}

{eq}\Longrightarrow f(x) = \biggr( \dfrac {5}{4^{3}} \biggr) \; \dfrac {1}{\biggr(1 + \dfrac {x}{4} \biggr)^{3}} {/eq}

We know the Binomial series expansion for the fractional negative powers:

{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 perfectly only when: {eq}\; \; \Longrightarrow |a| < 1 {/eq}

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

{eq}\displaystyle (1 + a)^{-3} = 1 - 3a + 6a^{2} - 10 a^{3} + \cdots {/eq}

Now replace the value of {eq}\; a = \biggr( \dfrac {x}{4} \biggr) \; {/eq} in the above expression:

{eq}\displaystyle \Biggr[1 + \biggr( \dfrac {x}{4} \biggr) \Biggr]^{-3} = 1 - \dfrac {3x}{4} + 6 \; \biggr( \dfrac {x}{4} \biggr)^{2} -10 \; \biggr( \dfrac {x}{4} \biggr)^{3} + \cdots {/eq}

The above series representation is valid only when: {eq}\; \; |\biggr( \dfrac {x}{4} \biggr)| < 1 \; \; \; \Longrightarrow \; \; |x| < 4 {/eq}

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \dfrac {5}{(4 + x)^{3}} = \biggr( \dfrac {5}{64} \biggr) \; \Biggr[ 1 - \dfrac {3x}{4} + 6 \; \biggr( \dfrac {x}{4} \biggr)^{2} -10 \; \biggr( \dfrac {x}{4} \biggr)^{3} + \cdots \Biggr]} {/eq}

{eq}\displaystyle \text {The radius of convergence is given as:} \; \; \Longrightarrow \; \; \text {R} = 4 {/eq}

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from Algebra II Textbook

Chapter 21 / Lesson 16