How do you use the binomial theorem to approximate #1.08^(1/2)# and hence find #sqrt(3)# to #4# significant figures?

Question

2706 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-11T07:16:23

Cutting the binomial series short,

#(1+x)^(1/2) ~= 1+x/2-x^2/8#

Then #sqrt(3) = 5/3sqrt(1.08) ~= 1+0.08/2-(0.08^2)/8 = 1.732#

#(1+x)^(1/2) = 1+(1/2)x+(1/(2!))(1/2)(1/2-1)x^2+(1/(3!))(1/2)(1/2-1)(1/2-2)x^3+...#

#=1+x/2-x^2/8+x^3/16-...#

If #x# is small this will converge rapidly

So with #x = 0.08# we get:

#sqrt(1.08) = (1+0.08)^(1/2) ~= 1 + 0.08/2 - (0.08^2)/8#

#=1+0.04-0.0008 = 1.0392#

Now #1.08 = 3^3/(5^2)#

So #sqrt(1.08) = sqrt(3^3/(5^2)) = sqrt(3*(3/5)^2) = 3/5sqrt(3)#

So #sqrt(3) = 5/3sqrt(1.08) ~= 5/3*1.0392 = 1.732#