Question

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

Answer 1

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 - (f(x_n))/(f'(x_n))`

Substituting for `f(x) = x^3 - 3` gives us:

`x_(n+1) = x_n - ((x_n)^3 - 3)/(3*(x_n)^2)`

Now, we pick an arbitrary number, (the closer it actually is to `root3(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 - ((0.5)^3 - 3)/(3*(0.5)^2) = 4.33333 bar 3`
`x_(2) = x_1 - ((x_1)^3 - 3)/(3*(x_1)^2) approx 2.94214333`
`x_(3) = x_2 - ((x_2)^3 - 3)/(3*(x_2)^2) approx 2.07695292`
`x_(4) = x_3 - ((x_3)^3 - 3)/(3*(x_3)^2) approx 1.61645303`
`x_(5) = x_4 - ((x_4)^3 - 3)/(3*(x_4)^2) approx 1.46034889`
`x_(6) = x_5 - ((x_5)^3 - 3)/(3*(x_5)^2) approx 1.44247296`
`x_(7) = x_6 - ((x_6)^3 - 3)/(3*(x_6)^2) approx 1.4422496`
`x_(8) = x_7 - ((x_7)^3 - 3)/(3*(x_7)^2) approx 1.44224957`

You can see that with only 8 iterations, we've obtained an approximation of `root 3(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.