How do you find the square root of 2000?

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Answer 1 · 2015-12-29T14:31:12

$sqrt(2000) = 20 sqrt(5) = 20 [2;bar(4)] ~~ 44.7$

If $a, b >= 0$ then $sqrt(ab) = sqrt(a)sqrt(b)$

So:

$sqrt(2000) = sqrt(400*5) = sqrt(400)*sqrt(5) = 20sqrt(5)$

Since $5 = 2^2+1$ is of the form $n^2+1$ , $sqrt(5)$ has a simple expansion as a continued fraction:

$sqrt(5) = [2;bar(4)] = 2 + 1/(4+1/(4+1/(4+1/(4+...))))$

According to how accurate an approximation we want we can terminate this continued fraction at more or fewer terms.

For example:

$sqrt(5) ~~ [2;4,4] = 2+1/(4+1/4) = 2 + 4/17 = 38/17$

So:

$sqrt(2000) = 20 sqrt(5) ~~ 20*38/17 ~~ 44.71$

Actually:

$sqrt(2000) ~~ 44.72135954999579392818$

As another way of calculating the successive approximations provided by the continued fraction, consider the sequence:

$0, 1, 4, 17, 72, 305,...$

where $a_1 = 0$ , $a_2 = 1$ , $a_(i+2) = a_i + 4a_(i+1)$

This is similar to the Fibonacci sequence, except the rule is $a_(i+2) = a_i + bb(4)a_(i+1)$ instead of $a_(i+2) = a_i + a_(i+1)$ .

This is strongly related to the continued fraction:

$[4;bar(4)] = 4+1/(4+1/(4+1/(4+1/(4+...))))$

The ratio between successive terms of the sequence tends to $2+sqrt(5)$ (somewhat faster than the Fibonacci sequence does to $1/2+sqrt(5)/2$ )

For example, we can find an approximation for $sqrt(5)$ in:

$305/72 - 2 = 161/72$

Hence $sqrt(2000) ~~ 20*161/72 = 3220/72 = 44.7dot(2)$