# How do you use Newton's Method to approximate the value of cube root?

Aug 22, 2014

The Newton-Raphson method approximates the roots of a function. So, we need a function whose root is the cube root we're trying to calculate.

Let's say we're trying to find the cube root of $3$. And let's say that $x$ is the cube root of $3$. Therefore,

${x}^{3} = 3$

For the Newton-Raphson method to be able to work its magic, we need to set this equation to zero.

${x}^{3} - 3 = 0$

Now we will recall the iterative equation for Newton-Raphson.

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

Substituting for $f \left(x\right) = {x}^{3} - 3$ gives us:

${x}_{n + 1} = {x}_{n} - \frac{{\left({x}_{n}\right)}^{3} - 3}{3 \cdot {\left({x}_{n}\right)}^{2}}$

Now, we pick an arbitrary number, (the closer it actually is to $\sqrt{3}$ the better) for ${x}_{0}$. Let's use ${x}_{0} = 0.5$. Then we substitute each previous number for ${x}_{n}$ back into the equation to get a closer and closer approximation to a solution of ${x}^{3} - 3 = 0$.

${x}_{1} = 0.5 - \frac{{\left(0.5\right)}^{3} - 3}{3 \cdot {\left(0.5\right)}^{2}} = 4.33333 \overline{3}$
${x}_{2} = {x}_{1} - \frac{{\left({x}_{1}\right)}^{3} - 3}{3 \cdot {\left({x}_{1}\right)}^{2}} \approx 2.94214333$
${x}_{3} = {x}_{2} - \frac{{\left({x}_{2}\right)}^{3} - 3}{3 \cdot {\left({x}_{2}\right)}^{2}} \approx 2.07695292$
${x}_{4} = {x}_{3} - \frac{{\left({x}_{3}\right)}^{3} - 3}{3 \cdot {\left({x}_{3}\right)}^{2}} \approx 1.61645303$
${x}_{5} = {x}_{4} - \frac{{\left({x}_{4}\right)}^{3} - 3}{3 \cdot {\left({x}_{4}\right)}^{2}} \approx 1.46034889$
${x}_{6} = {x}_{5} - \frac{{\left({x}_{5}\right)}^{3} - 3}{3 \cdot {\left({x}_{5}\right)}^{2}} \approx 1.44247296$
${x}_{7} = {x}_{6} - \frac{{\left({x}_{6}\right)}^{3} - 3}{3 \cdot {\left({x}_{6}\right)}^{2}} \approx 1.4422496$
${x}_{8} = {x}_{7} - \frac{{\left({x}_{7}\right)}^{3} - 3}{3 \cdot {\left({x}_{7}\right)}^{2}} \approx 1.44224957$

You can see that with only 8 iterations, we've obtained an approximation of $\sqrt{3}$ which is correct to 8 decimal places!

You can apply this same logic to whatever cube root you'd like to find, just use ${x}^{3} - a = 0$ as your equation instead, where $a$ is the number whose cube root you're looking for.