Question

How do you find a one-decimal place approximation for `sqrt 10`?

Answer 1

Use one step of a Newton Raphson method to find:

`sqrt(10) ~~ 19/6 ~~ 3.2`

Explanation:

To find the square root of a positive number `n`, choose a first approximation `a_0` then apply the following iteration step as many times as you like, to get the accuracy you want:

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

In our case, `n = 10` and let `a_0 = 3`, since `3^2 = 9` is quite close to what we want.

Then:

`a_1 = (a_0^2 + n)/(2a_0) = (3^2 + 10)/(2*3) = (9+10)/6 = 19/6 = 3.1dot(6)`

So to one decimal place `sqrt(10) ~~ 3.2`

Answer 2

Truncate the continued fraction expansion for `sqrt(10)` to find:

`sqrt(10) = [3;bar(6)] ~~ [3;6] = 3+1/6 = 3.1dot(6) ~~ 3.2`

Explanation:

The continued fraction expansion for `sqrt(n^2+1)` is

`sqrt(n^2+1) = [n;bar(2n)] = n + 1/(2n+1/(2n+1/(2n+1/(2n+...))))`

Now

`10 = 3^2 + 1`

So

`sqrt(10) = [3;bar(6)] = 3+1/(6+1/(6+1/(6+1/(6+...))))`

So truncating at `[3;6]` we find:

`sqrt(10) ~~ [3;6] = 3 + 1/6 = 3.1dot(6) ~~ 3.2`

If you want more accuracy you can include more terms, e.g.

`sqrt(10) ~~ [3;6,6,6] = 3+1/(6+1/(6+1/6)) = 3+37/228 ~~ 3.16228`