Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful numerical technique used to find the roots of a differentiable function $f(x) = 0$. The core idea is to start with an initial guess $x_0$ and iteratively refine it using the formula:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Setting up the Problem
To approximate $\sqrt[3]{2}$, we first need to define a function $f(x)$ whose root is equal to $\sqrt[3]{2}$.
If $x = \sqrt[3]{2}$, then cubing both sides gives: $$x^3 = 2 \implies x^3 - 2 = 0$$
Thus, we define our function as: $$f(x) = x^3 - 2$$
Next, we calculate the derivative: $$f'(x) = 3x^2$$
The Iterative Formula
Substituting our function and its derivative into the Newton-Raphson formula:
$$x_{n+1} = x_n - \frac{x_n^3 - 2}{3x_n^2}$$
Simplifying the expression: $$x_{n+1} = \frac{3x_n^3 - x_n^3 + 2}{3x_n^2} = \frac{2x_n^3 + 2}{3x_n^2} = \frac{2}{3} \left( x_n + \frac{1}{x_n^2} \right)$$
Step-by-Step Calculation
Let's choose an initial guess $x_0 = 1$ (since $1^3=1$ and $2^3=8$, the root is clearly between 1 and 2).
Iteration 1 ($x_1$): $x_1 = \frac{2(1)^3 + 2}{3(1)^2} = \frac{4}{3} \approx 1.333333$
Iteration 2 ($x_2$): $x_2 = \frac{2(1.333333)^3 + 2}{3(1.333333)^2} \approx 1.263889$
Iteration 3 ($x_3$): $x_3 = \frac{2(1.263889)^3 + 2}{3(1.263889)^2} \approx 1.259933$
Iteration 4 ($x_4$): $x_4 = \frac{2(1.259933)^3 + 2}{3(1.259933)^2} \approx 1.259921$
Verification
We continue iterating until the difference between consecutive approximations is less than $0.00001$. Comparing $x_3$ and $x_4$, the difference is roughly $0.000012$. One more iteration provides the required precision:
Final Result: $\sqrt[3]{2} \approx 1.25992$