Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful numerical technique used to find successively better approximations to the roots of a real-valued function $f(x) = 0$.
Given an initial guess $x_0$, the method uses the tangent line of the function at that point to find a closer approximation. The iterative formula is defined as:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Solving $x^3 - 18 = 0$
Given the function $f(x) = x^3 - 18$ on the interval $(2, 3)$, we follow these steps:
1. Define the Function and its Derivative
- $f(x) = x^3 - 18$
- $f'(x) = 3x^2$
2. Choose an Initial Guess
Since we are looking for the root in the interval $(2, 3)$, let's pick a midpoint as our initial guess:
- $x_0 = 2.5$
3. Iteration Process
Plug the formula into the iteration equation: $$x_{n+1} = x_n - \frac{x_n^3 - 18}{3x_n^2}$$
Iteration 1:
- $x_0 = 2.5$
- $f(2.5) = (2.5)^3 - 18 = 15.625 - 18 = -2.375$
- $f'(2.5) = 3(2.5)^2 = 18.75$
- $x_1 = 2.5 - (-2.375 / 18.75) \approx 2.5 + 0.1267 = 2.6267$
Iteration 2:
- $x_1 = 2.6267$
- $f(2.6267) = (2.6267)^3 - 18 \approx 18.134 - 18 = 0.134$
- $f'(2.6267) = 3(2.6267)^2 \approx 20.697$
- $x_2 = 2.6267 - (0.134 / 20.697) \approx 2.6267 - 0.0065 = 2.6202$
Intuition
After just two iterations, we have narrowed down the root to approximately 2.6202. The actual cube root of 18 is approximately 2.6207. You can see how rapidly this method converges to the true value by utilizing the gradient of the function to "aim" at the x-axis intercept.