Question

How do you find a one-decimal place approximation for `root5 26`?

Answer 1

Use one step of Newton's method to find `root(5)(26) ~~ 1.9`

Explanation:

To find an approximation for the `5`th root of a number `n`, choose a reasonable first approximation `a_0` then iterate using the following formula:

`a_(i+1) = a_i + (n - a_i^5)/(5 a_i^4)`

In our case let `n = 26` and `a_0 = 2` (since `2^5 = 32` is close)

Then:

`a_1 = a_0 + (n - a_0^5)/(5 a_0^4) = 2 + (26-32)/(5*16) = 2 - 6/80 = 2 - 0.075 = 1.925`

So `root(5)(26) ~~ 1.9`

If we want more accuracy, just apply the formula again to get `a_2` from `a_1`, etc.

This method is a particular case of Newton's method for finding a zero of a continuous differentiable function `f(x)`.

In our case `f(x) = x^5 - n`, `f'(x) = 5x^4` and Newton's method says to iterate using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i)) = a_i - (a_i^5 - n)/(5a_i^4) = a_i+(n-a_i^5)/(5a_i^4)`