How do you approximate #-sqrt30#?

2 Answers
MeneerNask
Jul 7, 2015

Let's forget the #-#sign for now. It will be put before the answer later. We do #sqrt30#.

Explanation:

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!

George C.
Jul 7, 2015

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#

Explanation:

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.

Related questions
See all questions in Simplification of Radical Expressions
Impact of this question
2516 views around the world
You can reuse this answer
Creative Commons License