Question

How do you calculate `root(3)(65)` to `4` decimal places ?

Answer 1

`root(3)(65) ~~ 4.0207`

Explanation:

Note that:

`65 = 4^3+1`

Method 1 - Newton Raphson

Let:

`f(x) = x^3-65`

Then:

`f'(x) = 3x^2`

So given an approximation `a_i` to `f(x) = 0` (i.e. to `root(3)(65)`), a better approximation is given by:

`a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (a_i^3-65)/(3a_i^2) = (2a_i^3+65)/(3a_i^2)`

Let `a_0 = 4`

Then:

`a_1 = (2a_0^3+65)/(3a_0^2) = (2(color(blue)(4))^3+65)/(3(color(blue)(4))^2) = 193/48 = 4.0208bar(3)`

This is almost correct to `4` decimal places, but let's iterate once more for greater accuracy...

`a_2 = (2a_1^3+65)/(3a_1^2) = (2(color(blue)(193/48))^3+65)/(3(color(blue)(193/48))^2) = (2(193)^3+65(48)^3)/(3(193)^2(48)) = 10783297/2681928 ~~ 4.02072576`

So to `4` decimal places `root(3)(65) ~~ 4.0207`

Method 2 - Binomial expansion

`root(3)(65) = 4 (1+1/64)^(1/3)`

`color(white)(root(3)(65)) = 4(1+1/3(1/64)-1/9(1/64)^2+5/81(1/64)^3-...)`

`color(white)(root(3)(65)) ~~ 4(1+1/192-1/36864)`

`color(white)(root(3)(65)) = 37055/9216`

`color(white)(root(3)(65)) ~~ 4.020725`

`color(white)(root(3)(65)) ~~ 4.0207" "` to `4` decimal places