Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful iterative technique used in numerical analysis to find the roots of a real-valued function $f(x) = 0$. The core idea is to start with an initial guess and repeatedly approximate the root using the tangent line to the function at that point.
The iterative formula is given by:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Where:
- $x_n$ is the current approximation.
- $x_{n+1}$ is the next, more accurate approximation.
- $f'(x_n)$ is the derivative of the function evaluated at $x_n$.
Solving the Problem
Problem: Find the positive root of $f(x) = x^3 - 2x - 5 = 0$ in the interval $(2, 3)$ correct to three decimal places.
Step 1: Define the function and its derivative
- $f(x) = x^3 - 2x - 5$
- $f'(x) = 3x^2 - 2$
Step 2: Choose an initial guess
Let's test $x=2$ and $x=3$:
- $f(2) = 2^3 - 2(2) - 5 = 8 - 4 - 5 = -1$
- $f(3) = 3^3 - 2(3) - 5 = 27 - 6 - 5 = 16$ Since $f(2)$ is closer to 0 than $f(3)$, we choose $x_0 = 2$ as our starting point.
Step 3: Perform Iterations
Iteration 1: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{-1}{3(2^2) - 2} = 2 - \frac{-1}{10} = 2.1$
Iteration 2: $f(2.1) = (2.1)^3 - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061$ $f'(2.1) = 3(2.1)^2 - 2 = 13.23 - 2 = 11.23$ $x_2 = 2.1 - \frac{0.061}{11.23} \approx 2.1 - 0.00543 = 2.09457$
Iteration 3: Using $x_2 = 2.09457$: $f(2.09457) = (2.09457)^3 - 2(2.09457) - 5 \approx 0.00028$ $f'(2.09457) = 3(2.09457)^2 - 2 \approx 11.162$ $x_3 = 2.09457 - \frac{0.00028}{11.162} \approx 2.094545$
Conclusion
Comparing $x_2$ and $x_3$, we see the values are consistent to three decimal places. Thus, the positive root of the equation correct to three decimal places is 2.095.