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

2 Answers
Oct 17, 2015

Answer:

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#

Oct 17, 2015

Answer:

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#