How do you find a one-decimal place approximation for #sqrt 37#?

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Answer 1 · 2015-10-18T15:08:37

Use one step of Newton Raphson method to find:

#sqrt(37) ~~ 73/12 = 6.08dot(3) ~~ 6.1#

To find the square root of a number #n#, choose a reasonable first approximation #a_0# and use the formula:

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

Repeat to get more accuracy.

For our purposes, #n = 37# and let #a_0 = 6# since #6^2=36#.

Then:

#a_1 = (a_0^2+n)/(2a_0) = (6^2+37)/(2*6) = (36+37)/12 = 73/12 = 6.08dot(3)#

We don't need any more steps to get the first decimal place, since we were pretty close to start.

Actually #sqrt(37)# is expressible as something called a continued fraction:

#sqrt(37) = [6;bar(12)] = 6+1/(12+1/(12+1/(12+...)))#

So you can also get approximations for #sqrt(37)# by just truncating this continued fraction and working out the value.

For example:

#sqrt(37) ~~ [6;12,12] = 6+1/(12+1/12) = 6 + 12/145 = 882/145 ~~ 6.08276#