Understanding the Newton-Raphson Method
The Newton-Raphson method is a powerful numerical technique used to find the roots of a differentiable function $f(x) = 0$. When we want to find the square root of a number $N$, we are essentially looking for the root of the equation:
$$f(x) = x^2 - N = 0$$
By applying the Newton-Raphson iterative formula, $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, we can derive a specific formula for square roots:
$$x_{n+1} = x_n - \frac{x_n^2 - N}{2x_n} = \frac{x_n^2 + N}{2x_n} = \frac{1}{2} \left( x_n + \frac{N}{x_n} \right)$$
Solving for √153
To find $\sqrt{153}$, we set $N = 153$. We need an initial guess $x_0$. Since $12^2 = 144$ and $13^2 = 169$, let's choose $x_0 = 12$ as a starting point.
Iteration 1:
Using $x_0 = 12$: $$x_1 = \frac{1}{2} \left( 12 + \frac{153}{12} \right) = \frac{1}{2} (12 + 12.75) = 12.375$$
Iteration 2:
Using $x_1 = 12.375$: $$x_2 = \frac{1}{2} \left( 12.375 + \frac{153}{12.375} \right)$$ $$x_2 = \frac{1}{2} (12.375 + 12.363636...) = 12.369318$$
Iteration 3:
Using $x_2 = 12.369318$: $$x_3 = \frac{1}{2} \left( 12.369318 + \frac{153}{12.369318} \right)$$ $$x_3 = \frac{1}{2} (12.369318 + 12.369318) = 12.369316$$
Conclusion
As we can see, the values have converged significantly by the third iteration. Rounding to three decimal places, we get:
$$\sqrt{153} \approx 12.369$$
This method is highly efficient, often doubling the number of correct decimal places with each iteration.