Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful numerical technique for finding the roots of a real-valued function $f(x) = 0$. It uses the tangent line at a current guess to approximate where the function hits the x-axis, providing a much closer estimate for the next iteration.
The iterative formula is given by:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Where:
- $x_n$ is the current guess.
- $f(x_n)$ is the function value at that guess.
- $f'(x_n)$ is the derivative of the function at that guess.
Solving $2x^2 - 3x - 1 = 0$
Given: $f(x) = 2x^2 - 3x - 1$ Derivative: $f'(x) = 4x - 3$ Initial guess: $x_0 = 1$ Stopping Criteria: $|x_{n+1} - x_n| < 10^{-4}$
Step 1: Iteration 1
$x_0 = 1$ $f(1) = 2(1)^2 - 3(1) - 1 = -2$ $f'(1) = 4(1) - 3 = 1$ $x_1 = 1 - (\frac{-2}{1}) = 1 + 2 = 3$
Step 2: Iteration 2
$x_1 = 3$ $f(3) = 2(3)^2 - 3(3) - 1 = 18 - 9 - 1 = 8$ $f'(3) = 4(3) - 3 = 9$ $x_2 = 3 - (\frac{8}{9}) \approx 3 - 0.8889 = 2.1111$
Step 3: Iteration 3
$x_2 = 2.1111$ $f(2.1111) = 2(2.1111)^2 - 3(2.1111) - 1 \approx 1.8025$ $f'(2.1111) = 4(2.1111) - 3 = 5.4444$ $x_3 = 2.1111 - (\frac{1.8025}{5.4444}) \approx 2.1111 - 0.3311 = 1.7800$
Step 4: Iteration 4
$x_3 = 1.7800$ $f(1.7800) = 2(1.7800)^2 - 3(1.7800) - 1 \approx 0.2528$ $f'(1.7800) = 4(1.7800) - 3 = 4.1200$ $x_4 = 1.7800 - (\frac{0.2528}{4.1200}) \approx 1.7800 - 0.0614 = 1.7186$
Step 5: Iteration 5
$x_4 = 1.7186$ $f(1.7186) \approx 0.0076$ $f'(1.7186) \approx 3.8744$ $x_5 = 1.7186 - (\frac{0.0076}{3.8744}) \approx 1.7166$
As we continue, the value of $x_n$ will converge toward the root $\frac{3 + \sqrt{17}}{4} \approx 1.78077$ or the other root $\frac{3 - \sqrt{17}}{4} \approx -0.28077$ depending on the initial guess. In this case, starting at $x_0=1$ leads us toward the positive root.