Question

How do you find a one-decimal place approximation for `-root4 200`?

Answer 1

Use some reasonings about `3^4`, `4^4` and `sqrt(200)` to choose approximation `3.8`, then use one step of Newton's method to refine the approximation, finding that `3.8` is good to one decimal place already.

Explanation:

`3^4 = 81` and `4^4 = 256`, so `root(4)(200)` lies somewhere between `3` and `4`.

More specifically, `200 ~~ 196 = 14^2` so `root(4)(200) ~~ sqrt(14)` and `sqrt(14) ~~ 4-1/4 = 3.75`, so `root(4)(200) ~~ 3.8`

To find the `4`th root of a number `n`, we could find its square root and then the square root of that, but instead let's choose a reasonable first approximation `a_0`, then use the following formula to iterate:

`a_(i+1) = a_i + (n - a_i^4)/(4a_i^3)`

Let `a_0 = 3.8`

Then:

`a_1 = a_0 + (200-3.8^4)/(4*3.8^3)`

`=3.8 + (200-208.5136)/219.488`

`=3.8 - 8.5136/219.488 ~~ 3.76`

So `3.8` is a good approximation of `root(4)(200)` to one decimal place and `-3.8` is a one decimal place approximation for `-root(4)(200)`