How do you find a one-decimal place approximation for #root3 30#?

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Answer 1 · 2015-10-18T12:33:05

Use one step of Newton's method to find #root(3)(30) ~~ 3.1#

To find the cube root of a number #n#, start with an approximation #a_0# and apply the following iteration step:

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

This is based on Newton's method for finding the zero of a function #f(x)# using:

#a_(i+1) = a_i - f(x)/(f'(x))#

In our case #f(x) = x^3 - n# and #f'(x) = 3x^2#

Since #3^3 = 27# is not far off, let us use #a_0 = 3#.

Then:

#a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 3+(30-27)/(3*3^2) = 3+3/27 = 3+1/9 = 3.dot(1)#

So to one decimal place #root(3)(30) ~~ 3.1#