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)#