How do you find a one-decimal place approximation for #sqrt 18#?

To find an approximation for the square root of a number #n#, start with a reasonable approximation #a_0# and apply the following formula to make a better approximation. Repeat to get better approximations.

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

In our case #n = 18# and it makes sense to choose #a_0 = 4# since #4^2 = 16# is fairly close.

Then:

#a_1 = (a_0^2+n)/(2a_0) = (4^2+18)/(2*4) = (16+18)/8 = 34/8 = 17/4 = 4.25#

Unfortunately #a_1# is right in the middle between #4.2# and #4.3#, so let's try another iteration...

#a_2 = (a_1^2+n)/(2a_1) = ((17/4)^2+18)/(2*(17/4))#

#=(289/16+18)/(17/2) =(289+18*16)/(17*8)=(289+288)/136#

#=577/136 ~~ 4.24265#

That's more decimal places than we need, but at least we can say with confidence that #sqrt(18) ~~ 4.2# to one decimal place.

Alternatively, you can use the method described in http://socratic.org/questions/given-an-integer-n-is-there-an-efficient-way-to-find-integers-p-q-such-that-abs-
to find a continued fraction expansion for #sqrt(18)# ...

#sqrt(18) = [4;bar(4,8)] = 4+1/(4+1/(8+1/(4+1/(8+1/(4+...)))))#

Then you can approximate #sqrt(18)# by truncating the continued fraction.

For example:

#sqrt(18) ~~ [4;4,8] = 4+1/(4+1/8) = 4+8/33 = 140/33 = 4.dot(2)dot(4)#