# Using a binomial series find an approximate value to 7^{1 / 3}. If x is your answer, the x must...

## Question:

Using a binomial series find an approximate value to {eq}7^{\dfrac 1 3} {/eq}. If x is your answer, the x must satisfy absolute value of {eq}(7^{\dfrac 1 3} - x) {/eq} is less than {eq}10^{-4} {/eq}.

## Binomial Series Representation:

{eq}\\ {/eq}

The Binomial series representation for the fractional power is defined as:

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

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

Here we will assume a function {eq}\; f(x) \; {/eq} which help us in order to approximate the value of {eq}\; \sqrt [3] {7} {/eq}. First of all, we will determine the series representation of the assumed function with the help of above expression then we will put some value of {eq}\; x \; {/eq} that will give us the approximation.

{eq}\\ {/eq}

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

{eq}\displaystyle f(x) = \sqrt [3]{6} \; \Biggr[1 + \biggr( \dfrac {x}{6} \biggr) \Biggr]^{\dfrac {1}{3}} {/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 {a}{3} - \dfrac {a^{2}}{9} + \dfrac {5a^{3}}{81} + \cdots {/eq}

Now replace the value of {eq}\; a = \biggr( \dfrac {x}{6} \biggr) \; {/eq} in the above expression:

{eq}\displaystyle \Biggr[1 + \biggr( \dfrac {x}{6} \biggr) \Biggr]^{\dfrac {1}{3}} = 1 + \dfrac {1}{3} \; \biggr( \dfrac {x}{6} \biggr) - \dfrac {1}{9} \; \biggr( \dfrac {x}{6} \biggr)^{2} + \dfrac {5}{81} \; \biggr( \dfrac {x}{6} \biggr)^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; |\dfrac {x}{6}| < 1 \; \; \; \Longrightarrow \; |x| < 6 {/eq}

{eq}\displaystyle \sqrt [3] {6 + x} = \sqrt [3] {6} \; \Biggr[ 1 + \dfrac {1}{3} \; \biggr( \dfrac {x}{6} \biggr) - \dfrac {1}{9} \; \biggr( \dfrac {x}{6} \biggr)^{2} + \dfrac {5}{81} \; \biggr( \dfrac {x}{6} \biggr)^{3} + \cdots \Biggr] {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |x| < 6 {/eq}

Now replace the value of {eq}\; x = 1 \; {/eq} in the above expression in order to approximate the value of {eq}\; \sqrt [3] {7} \; {/eq}:

{eq}\displaystyle \sqrt [3] {7} \approx \sqrt [3] {6} \; \Biggr[ 1 + \dfrac {1}{3 \times 6} - \dfrac {1}{9 \times 6^{2}} + \dfrac {5}{81 \times 6^{3}} \Biggr] \; \approx \; 1.91298 {/eq}

{eq}\displaystyle \Longrightarrow \boxed {\sqrt [3] {7} \; \approx \; 1.91298} {/eq}