How do you find a one-decimal place approximation for #root3 15#?

1 Answer
Oct 23, 2015

Answer:

Find #2 < root(3)(15) < 3#, then using #2.5# as a first approximation and one step of Newton's method, find that this is already good to one decimal place.

Explanation:

Note that #2^3 = 8 < 15 < 27 = 3^3#

So #2 < root(3)(15) < 3#

To find an approximation for the cube root of a number #n#, choose a reasonable first approximation #a_0# and apply the following formula (from Newton's method), repeatedly if necessary:

#a_(i+1) = a_i + (n - a_i^3)/(3a_i^2)#

In our case #n = 15# and choose #a_0 = 2.5# since the cube root is somewhere between #2# and #3#.

Then:

#a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 2.5 + (15 - 2.5^3)/(3*2.5^2)#

#=2.5 + (15-15.625)/18.75 = 2.5 - 0.625/18.75 = 2.4dot(6)dot(6)#

So our first approximation was already good to one decimal place.