Question

How do you approximate `-sqrt30`?

Answer 1

Let's forget the `-`sign for now. It will be put before the answer later. We do `sqrt30`.

Explanation:

Now `25` is the nearest square below `30`, and `36` is the first higher square.
So the answer must be between `5and6`.

Since `30` is also roughly halfway between `25and36`, the answer should also be in that neighbourhood.
My first approximation would be `5.5` which gives a square of `30.25`, a little bit high.

The calculator answer is `5.477...` which will be rounded to `5.5`
Not bad!

Answer 2

Use Newton Raphson method to find a sequence of approximations for `sqrt(30)`:

`11/2 = 5.5`

`241/44 ~= 5.4772`

`116161/21208 ~= 5.477225575`

Explanation:

The Newton-Raphson method, specialised to finding square roots boils down to choosing a reasonable first approximation `a_0`, then iterating using the formula:

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

where `n` is the number you are approximating the square root of.

In our case let us choose `a_0 = 5`, `n = 30`.

Then:

`a_1 = (a_0^2+n)/(2a_0)`

`=(5^2+30)/(2*5) = (25+30)/10 = 55/10 = 11/2 = color(red)(5.5)`

`a_2 = (a_1^2+n)/(2a_1)`

`=((11/2)^2+30)/(2*11/2) =(121/4+120/4)/11=241/44 ~= color(red)(5.4772)`

`a_3 = (a_2^2+n)/(2a_2)`

`=((241/44)^2+30)/(2*241/44)`

`=(58081/(44^2)+58080/(44^2))/(482/44)`

`=116161/21208 ~= color(red)(5.477225575)`

Notice that the number of significant digits roughly doubles on every iteration.

Actually, if we want to use rational approximations,

it is nicer to write `a_0 = p_0/q_0` and iterate using the formulas:

`p_(i+1) = p_i^2+q_i^2n`

`q_(i+1) = 2p_iq_i`

optionally dividing `p_i` and `q_i` by any common factor.

That avoids some of the messy fractions.