Question

How do you approximate `sqrt(17.02)`?

Answer 1

`sqrt(17.02)~~color(green)4.125508`

Explanation:

Possibly NOT the intended method since this was asked under "Using Newton's Method"
We could take `4` as a really gross estimate of `sqrt(17.02)` since `4^2=16` is closer to `17.02` than `5^2=25`

We know that `4` really is too small though,
so lets suppose our answer is `4+epsilon` (with `abs(epsilon) < 1`)
`color(white)("XXX")(4+epsilon)^2=17.02`

`color(white)("XXX")16+8epsilon+epsilon^2=17.02`

`color(white)("XXX")8epsilon =1.02-epsilon^2`

`color(white)("XXX")epsilon = 1.02/8-epsilon^2/8`

Since `abs(epsilon) < 1` then `epsilon^2` won't be very big and `epsilon^2/8` will be even smaller.
So as an initial estimate for `epsilon` we could try using
`color(white)("XXX")epsilon' = 1.01/8=0.12625`
in place of `epsilon` on the right side of this equation.

`color(white)("XXX")epsilon ~~(1.02-(0.12625)^2)/8`

(I used a calculator from this point on)

`color(white)("XXX")epsilon ~~0.125508`

`4+epsilon = 4.125508`

Testing this value as the root of `17.02`
`color(white)("XXX")4.125508^2~~17.01981`
which (to my mind) is close enough

Answer 2

`sqrt(17.02)~~4.125531` (after 20 iterations using the Newton Method)

Explanation:

If you wanted the answer using Newton's Method; here it is
as the product of some iterative expressions in a spreadsheet:

Answer 3

4.1255302699 to 9 Decimal Places

Explanation:

`sqrt(17.02)` is a solution to the equation
`x^2=17.02 => x^2-17.02 =0`

Let `f(x)=x^2-17.02` then `f'(x)=2x` and we can use Newton's method using the iterative formula;

`x_n = x_(n-1) - f(x_(n-1)) / (f'(x_(n-1)))`

If we start with `x_0=4`, then we can tabulate the results as follows (in this case using Excel);

So we see that very rapidly the method converges to a solution 4.1255302699 to 9 Decimal Places