The problem involves an iterative formula:

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

Starting with $x_1 = 10$, we are tasked to calculate $x_3$ and $x_4$, rounding the results to 1 decimal place.

Step-by-step solution:

1. Start with $x_1 = 10$: This is the given initial value.

2. Calculate $x_2$: Using the formula: $x_2 = \frac{x_1}{2} + \frac{1}{x_1}$ Substituting $x_1 = 10$: $x_2 = \frac{10}{2} + \frac{1}{10} = 5 + 0.1 = 5.1$

3. Calculate $x_3$: Using the formula: $x_3 = \frac{x_2}{2} + \frac{1}{x_2}$ Substituting $x_2 = 5.1$: $x_3 = \frac{5.1}{2} + \frac{1}{5.1} = 2.55 + 0.196 = 2.746$ Rounding to 1 decimal place: $x_3 = 2.7$

4. Calculate $x_4$: Using the formula: $x_4 = \frac{x_3}{2} + \frac{1}{x_3}$ Substituting $x_3 = 2.7$: $x_4 = \frac{2.7}{2} + \frac{1}{2.7} = 1.35 + 0.370 = 1.72$ Rounding to 1 decimal place: $x_4 = 1.7$

Final answers:

• $x_3 = 2.7$
• $x_4 = 1.7$

Would you like further clarification or assistance with related problems?

Related Questions:

1. How does the iterative method converge for different starting values of $x_1$?
2. What is the mathematical significance of this iterative formula?
3. Can you generalize the process to find $x_n$ for any $n$?
4. What happens if $x_1$ is a very small number like $0.1$?
5. How can this iteration be used to approximate square roots?

Tip:

When solving iterative formulas, always keep intermediate calculations accurate before rounding to ensure precise results in subsequent steps!