This question asks us to calculate the second iterate ($x_2$) of a given recursive formula to estimate $\sqrt{10}$, starting from an initial guess $x_0 = 3.1$. The recursive formula provided is:

$x_n = \frac{x_{n-1} + \frac{10}{x_{n-1}}}{2}.$

Let's compute $x_1$ and then $x_2$ step by step.

Step 1: Calculate $x_1$

Substitute $x_0 = 3.1$ into the formula: $x_1 = \frac{3.1 + \frac{10}{3.1}}{2}.$

First, compute $\frac{10}{3.1}$: $\frac{10}{3.1} \approx 3.22581.$

Now add $3.1$ to $3.22581$: $3.1 + 3.22581 = 6.32581.$

Divide by 2: $x_1 = \frac{6.32581}{2} \approx 3.16291.$

Step 2: Calculate $x_2$

Use $x_1 \approx 3.16291$ in the formula: $x_2 = \frac{3.16291 + \frac{10}{3.16291}}{2}.$

First, compute $\frac{10}{3.16291}$: $\frac{10}{3.16291} \approx 3.16202.$

Now add $3.16291$ to $3.16202$: $3.16291 + 3.16202 = 6.32493.$

Divide by 2: $x_2 = \frac{6.32493}{2} \approx 3.16227.$

Final Answer:

The second iterate is approximately: $\boxed{x_2 = 3.16227}.$

Would you like further explanation or verification?

Here are 5 related questions for further exploration:

1. How does the recursive formula for estimating square roots converge to the actual value?
2. Why is it important to carry out calculations to multiple decimal places in iterative methods?
3. Can this formula be used to estimate $\sqrt{20}$? How would it differ?
4. What is the significance of the initial guess ($x_0$) in the speed of convergence?
5. How would the error between iterations be calculated in this case?

Tip: Iterative methods often improve accuracy significantly with each step, but watch for computational round-off errors in practical applications!