Question

How do you find a one-decimal place approximation for `root3 15`?

Answer 1

Find `2 < root(3)(15) < 3`, then using `2.5` as a first approximation and one step of Newton's method, find that this is already good to one decimal place.

Explanation:

Note that `2^3 = 8 < 15 < 27 = 3^3`

So `2 < root(3)(15) < 3`

To find an approximation for the cube root of a number `n`, choose a reasonable first approximation `a_0` and apply the following formula (from Newton's method), repeatedly if necessary:

`a_(i+1) = a_i + (n - a_i^3)/(3a_i^2)`

In our case `n = 15` and choose `a_0 = 2.5` since the cube root is somewhere between `2` and `3`.

Then:

`a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 2.5 + (15 - 2.5^3)/(3*2.5^2)`

`=2.5 + (15-15.625)/18.75 = 2.5 - 0.625/18.75 = 2.4dot(6)dot(6)`

So our first approximation was already good to one decimal place.