The problem involves using Newton's Method to approximate a zero of the function $f(x) = \cos(x)$, with an initial guess of $x_1 = 1.3$.

Newton's Method formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Steps to solve:

1. Calculate $f(x)$ and $f'(x)$:

   • $f(x) = \cos(x)$
   • $f'(x) = -\sin(x)$ (the derivative of $\cos(x)$)

2. Start with the given initial guess $x_1 = 1.3$.

Let's go step by step through the first and second iterations.

Iteration 1 (for $n = 1$):

• $x_1 = 1.3$
• $f(x_1) = \cos(1.3)$
• $f'(x_1) = -\sin(1.3)$

Using a calculator:

• $\cos(1.3) \approx 0.2675$
• $-\sin(1.3) \approx -0.9636$

Now calculate the Newton's Method step: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.3 - \frac{0.2675}{-0.9636} \approx 1.5776$

Iteration 2 (for $n = 2$):

• $x_2 \approx 1.5776$
• $f(x_2) = \cos(1.5776)$
• $f'(x_2) = -\sin(1.5776)$

Using a calculator:

• $\cos(1.5776) \approx -0.0002$
• $-\sin(1.5776) \approx -0.9999$

Now calculate the next Newton's Method step: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.5776 - \frac{-0.0002}{-0.9999} \approx 1.5778$

Filling in the table:

Would you like more details on how the values were calculated, or have any other questions? Here are some related questions:

1. What is the significance of Newton's Method in solving equations?
2. How does the choice of initial guess affect the convergence of Newton's Method?
3. Can Newton's Method fail to converge, and if so, under what conditions?
4. Why do we need to compute the derivative in Newton's Method?
5. How accurate is Newton's Method compared to other root-finding algorithms?

Tip: Newton's Method converges faster when the initial guess is closer to the actual root.