To find the square root of 2 using the Newton-Raphson method to 2 decimal places, follow these steps:

The Newton-Raphson iteration formula is:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Here, we are solving $f(x) = x^2 - 2 = 0$, so:

• $f(x) = x^2 - 2$
• $f'(x) = 2x$

Steps:

1. Initial guess: Choose an initial guess close to the square root of 2, say $x_0 = 1.5$.
2. Iteration: Compute $x_{n+1}$ using the formula: $x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}$
3. Stop when the change between successive iterations is less than 0.01 (two decimal accuracy).

Iteration 1:

$x_1 = x_0 - \frac{x_0^2 - 2}{2x_0} = 1.5 - \frac{1.5^2 - 2}{2 \cdot 1.5}$ $x_1 = 1.5 - \frac{2.25 - 2}{3} = 1.5 - \frac{0.25}{3} = 1.5 - 0.0833 = 1.4167$

Iteration 2:

$x_2 = x_1 - \frac{x_1^2 - 2}{2x_1} = 1.4167 - \frac{1.4167^2 - 2}{2 \cdot 1.4167}$ $x_2 = 1.4167 - \frac{2.0069 - 2}{2.8334} = 1.4167 - \frac{0.0069}{2.8334} = 1.4167 - 0.0024 = 1.4143$

Iteration 3:

$x_3 = x_2 - \frac{x_2^2 - 2}{2x_2} = 1.4143 - \frac{1.4143^2 - 2}{2 \cdot 1.4143}$ $x_3 = 1.4143 - \frac{2.00000649 - 2}{2.8286} = 1.4143 - \frac{0.00000649}{2.8286} = 1.4143 - 0.0000023 \approx 1.4142$

Result:

The square root of 2 to 2 decimal places is approximately:

$\boxed{1.41}$

Would you like a more detailed explanation of any step?

---

Here are 5 related questions to expand on this:

1. How does the Newton-Raphson method ensure convergence?
2. What are other root-finding methods similar to Newton-Raphson?
3. Can Newton-Raphson fail to converge, and under what conditions?
4. How would you modify the Newton-Raphson method for functions with multiple roots?
5. What is the general geometric interpretation of the Newton-Raphson iteration?

Tip: Always ensure your initial guess is reasonably close to the root for faster and reliable convergence.