Question

How do you find a one-decimal place approximation for `-root3 35`?

Answer 1

Consider `(3+t)^3` to find `-root(3)(35)~~-3.3`

Explanation:

We know that `3^3=27`, so consider `(3+t)^3`...

`(3+t)^3=3^3+3*3^2t+3*3t^2+t^3`

`=27+27t+9t^2+t^3`

`~~27+27t` for small values of `t`

So approximating...

`35=27+27t`

Hence

`t=(35-27)/27=8/27~~0.3`

So `root(3)(35)~~3+0.3=3.3` and `-root(3)(35)~~-3.3`

More generally, to find the cube root of a number `n`, starting with an approximation `a_0`, iterate using the formula:

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

This is a form of Newton Raphson method. Given a function `f(x)` for which we want to solve `f(x) = 0`, start with an approximation `a_0` and iterate using:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

In our case `f(x) = x^3 - n` and `f'(x) = 3x^2`