How do you approximate #sqrt5#?

2 Answers
Jul 5, 2015

Answer:

#sqrt(5)~~2.2#

Explanation:

#2^2 = 4# and #3^2= 9#
So #2 < sqrt(5) < 3#

Midpoint between 2 and 3 is 2.5
#color(white)("XXXX")##2.5^2 = ((25)/10)^2 = 625/100 = 6.25#
So #2 < sqrt(5) < 2.5#

Midpoint between 2 and 2.5 is (approximately) 2.2
#color(white)("XXXX")##2.2^2 = ((22)/10)^2 = 484/100 = 4.84#
So #2.2 < sqrt(5) < 2.5#

Midpoint between 2.2 and 2.5 is (approximately) 2.3
#color(white)("XXXX")##2.3^2 = ((23)/10)^2 = 529/100 = 5.29#
So #2.2 < sqrt(5) < 2.3#

Obviously you could care this process on getting a smaller range each time.

The process is basically the Newton Method. For each cycle find the midpoint of the range, evaluate the square at that value, and replace either the upper or lower bound of the range. If you need to do the process manually, I've found that working with decimal fractions (and converting to fractions with a denominator which is a power of 10) as demonstrated, reduces the effort.

Jul 5, 2015

Answer:

Choose a reasonable first approximation #a_0#, then iterate using the formula:

#a_(i+1) = (a_i^2+5)/(2a_i)#

If #a_0 = 2# then #a_2 ~= 2.236# which is pretty good.

Explanation:

Let #a_0 = 2#

Then using the formula:

#a_1 = (a_0^2+5)/(2a_0)#

#=(5+2^2)/(2*2)#

#=9/4 = 2.25#

Iterate again:

#a_2 = (a_1^2+5)/(2a_1)#

#=((9/4)^2+5)/(2*9/4)#

#=(81/16+80/16)/(9/2)#

#=(161/16)/(9/2)#

#=161/72 = 2.236dot(1)#

This is certainly good enough for 4 significant digits.

#sqrt(5) ~= 2.2360679775#