How do you approximate #-sqrt30#?
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Let's forget the #-#sign for now. It will be put before the answer later. We do #sqrt30#.
Now #25# is the nearest square below #30#, and #36# is the first higher square.
So the answer must be between #5and6#.
Since #30# is also roughly halfway between #25and36#, the answer should also be in that neighbourhood.
My first approximation would be #5.5# which gives a square of #30.25#, a little bit high.
The calculator answer is #5.477...# which will be rounded to #5.5#
Not bad!
Use Newton Raphson method to find a sequence of approximations for #sqrt(30)#:
#11/2 = 5.5#
#241/44 ~= 5.4772#
#116161/21208 ~= 5.477225575#
The Newton-Raphson method, specialised to finding square roots boils down to choosing a reasonable first approximation #a_0#, then iterating using the formula:
#a_(i+1) = (a_i^2+n)/(2a_i)#
where #n# is the number you are approximating the square root of.
In our case let us choose #a_0 = 5#, #n = 30#.
Then:
#a_1 = (a_0^2+n)/(2a_0)#
#=(5^2+30)/(2*5) = (25+30)/10 = 55/10 = 11/2 = color(red)(5.5)#
#a_2 = (a_1^2+n)/(2a_1)#
#=((11/2)^2+30)/(2*11/2) =(121/4+120/4)/11=241/44 ~= color(red)(5.4772)#
#a_3 = (a_2^2+n)/(2a_2)#
#=((241/44)^2+30)/(2*241/44)#
#=(58081/(44^2)+58080/(44^2))/(482/44)#
#=116161/21208 ~= color(red)(5.477225575)#
Notice that the number of significant digits roughly doubles on every iteration.
Actually, if we want to use rational approximations,
it is nicer to write #a_0 = p_0/q_0# and iterate using the formulas:
#p_(i+1) = p_i^2+q_i^2n#
#q_(i+1) = 2p_iq_i#
optionally dividing #p_i# and #q_i# by any common factor.
That avoids some of the messy fractions.
