How do you use Newton's Method to approximate the value of cube root?

Question

Charles Anctil · Accepted Answer

The Newton-Raphson method approximates the roots of a function. So, we need a function whose root is the cube root we're trying to calculate. Let's say we're trying to find the cube root of #3#. And let's say that #x# is the cube root of #3#. Therefore, #x^3 = 3# For the Newton-Raphson method to be able to work its magic, we need to set this equation to zero. #x^3 - 3 = 0# Now we will recall the iterative equation for Newton-Raphson. #x_(n+1) = x_n - (f(x_n))/(f'(x_n))# Substituting for #f(x) = x^3 - 3# gives us: #x_(n+1) = x_n - ((x_n)^3 - 3)/(3*(x_n)^2)# Now, we pick an arbitrary number, (the closer it actually is to #root3(3)# the better) for #x_0#. Let's use #x_0 = 0.5#. Then we substitute each previous number for #x_n# back into the equation to get a closer and closer approximation to a solution of #x^3 - 3 = 0#. #x_(1) = 0.5 - ((0.5)^3 - 3)/(3*(0.5)^2) = 4.33333 bar 3#
#x_(2) = x_1 - ((x_1)^3 - 3)/(3*(x_1)^2) approx 2.94214333#
#x_(3) = x_2 - ((x_2)^3 - 3)/(3*(x_2)^2) approx 2.07695292#
#x_(4) = x_3 - ((x_3)^3 - 3)/(3*(x_3)^2) approx 1.61645303#
#x_(5) = x_4 - ((x_4)^3 - 3)/(3*(x_4)^2) approx 1.46034889#
#x_(6) = x_5 - ((x_5)^3 - 3)/(3*(x_5)^2) approx 1.44247296#
#x_(7) = x_6 - ((x_6)^3 - 3)/(3*(x_6)^2) approx 1.4422496#
#x_(8) = x_7 - ((x_7)^3 - 3)/(3*(x_7)^2) approx 1.44224957# You can see that with only 8 iterations, we've obtained an approximation of #root 3(3)# which is correct to 8 decimal places! You can apply this same logic to whatever cube root you'd like to find, just use #x^3 - a = 0# as your equation instead, where #a# is the number whose cube root you're looking for.

Isabella Ackerman · Answer

undefined To use Newton's Method to approximate the value of a cube root, follow these steps:

1. Choose a starting guess $ x_0 $ for the cube root.
2. Use the formula $ x_{n+1} = x_n - \frac{{f(x_n)}}{{f'(x_n)}} $ to iterate and refine the approximation, where $ f(x) = x^3 - a $ is the function whose root we want to find (where $ a $ is the number whose cube root we're approximating) and $ f'(x) $ is the derivative of $ f(x) $.
3. Repeat step 2 until the difference between successive approximations $ |x_{n+1} - x_n| $ is smaller than a chosen tolerance level or until a desired level of precision is achieved.

This iterative process converges towards the value of the cube root of $ a $.