Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful iterative technique used in numerical analysis to find roots of a real-valued function $f(x) = 0$. It uses the concept of tangent lines to approximate the root of a function, moving closer to the true value with each iteration.
The Formula
Given an initial guess $x_0$, the next approximation $x_{n+1}$ is calculated using the formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Solving $x^3 + x - 1 = 0$
We are looking for a solution for $f(x) = x^3 + x - 1$ in the interval $[0, 1]$ accurate to $10^{-4}$.
Step 1: Define the function and its derivative
- $f(x) = x^3 + x - 1$
- $f'(x) = 3x^2 + 1$
Step 2: Choose an initial guess ($x_0$)
Let's test the bounds of the interval:
- $f(0) = 0^3 + 0 - 1 = -1$
- $f(1) = 1^3 + 1 - 1 = 1$ Since the function changes sign between $0$ and $1$, a root exists in $[0, 1]$. We can pick $x_0 = 0.5$ as our starting point.
Step 3: Iterations
Iteration 1:
- $f(0.5) = (0.5)^3 + 0.5 - 1 = 0.125 + 0.5 - 1 = -0.375$
- $f'(0.5) = 3(0.5)^2 + 1 = 0.75 + 1 = 1.75$
- $x_1 = 0.5 - (-0.375 / 1.75) \approx 0.7143$
Iteration 2:
- $f(0.7143) = (0.7143)^3 + 0.7143 - 1 \approx 0.0787$
- $f'(0.7143) = 3(0.7143)^2 + 1 \approx 2.5307$
- $x_2 = 0.7143 - (0.0787 / 2.5307) \approx 0.6832$
Iteration 3:
- $f(0.6832) = (0.6832)^3 + 0.6832 - 1 \approx 0.0019$
- $f'(0.6832) = 3(0.6832)^2 + 1 \approx 2.4005$
- $x_3 = 0.6832 - (0.0019 / 2.4005) \approx 0.6823$
Conclusion
Performing one more iteration, we find the value stabilizes. The root of the equation $x^3 + x - 1 = 0$ is approximately 0.6823.