give answer in 2 step with explanation at the end of each step and final answer at the end:Problem 4 (20 points)
The figure below depicts the generating function g(x) used in a Simple Fixed-Point ite

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Problem 4 (20 points)
The figure below depicts the generating function g(x) used in a Simple Fixed-Point iteration algorithm
(and a 45° line for reference).
(a) (6 points) Starting at a value of x0 = 3, give x1, x2, x3, and x4 generated by the fixed-point method.
Show your work directly on the plot below.
(b) (8 points) Give the values of the approximate percent relative error, Es in the first 4 steps of the
fixed-point algorithm, based on your values from part (a)
(9) (6 points) Is the method converging or diverging? Do the graphical results agree with the theory
covered in the course (Fixed Point Theorem). Please discuss in the space below.
Simple Fixed-Point
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VIPSolutions · Accepted Answer

Let's solve the problem step by step as requested.

---
### Step 1: Find the Fixed-Point Iterations (x₁, x₂, x₃, x₄)

#### **Explanation:**
The fixed-point iteration uses the formula:  
$$ x_{n+1} = g(x_n) $$

- Start at $ x_0 = 3 $ (marked on the plot).
- To find $ x_1 $, read $ g(3) $ from the red curve.
- Then, $ x_2 = g(x_1) $, and so on.

**From the plot:**
- At $ x_0 = 3 $, $ g(x_0) \approx 0.5 $ (where the red curve is at $ x = 3 $).
  - So, $ x_1 \approx 0.5 $
- At $ x_1 \approx 0.5 $, $ g(x_1) \approx 2.2 $
  - So, $ x_2 \approx 2.2 $
- At $ x_2 \approx 2.2 $, $ g(x_2) \approx 1.1 $
  - So, $ x_3 \approx 1.1 $
- At $ x_3 \approx 1.1 $, $ g(x_3) \approx 1.8 $
  - So, $ x_4 \approx 1.8 $

**Summary Table:**
| iter |   x   |
|:----:|:-----:|
|  1   | 0.5   |
|  2   | 2.2   |
|  3   | 1.1   |
|  4   | 1.8   |

#### **Explanation:**
- Each iteration is calculated by plugging the previous value into the function $ g(x) $ as shown on the plot.

---

### Step 2: Calculate the Approximate Percent Relative Error $ E_a $

Formula for approximate percent relative error:
$$
E_a = \left| \frac{x_{n+1} - x_n}{x_{n+1}} \right| \times 100\%
$$

Let's compute for each step:

#### Iteration 1 (from x₀ to x₁):
$$
E_{a,1} = \left| \frac{0.5 - 3}{0.5} \right| \times 100\% = \left| \frac{-2.5}{0.5} \right| \times 100\% = 500\%
$$

#### Iteration 2 (from x₁ to x₂):
$$
E_{a,2} = \left| \frac{2.2 - 0.5}{2.2} \right| \times 100\% = \left| \frac{1.7}{2.2} \right| \times 100\% \approx 77.3\%
$$

#### Iteration 3 (from x₂ to x₃):
$$
E_{a,3} = \left| \frac{1.1 - 2.2}{1.1} \right| \times 100\% = \left| \frac{-1.1}{1.1} \right| \times 100\% = 100\%
$$

#### Iteration 4 (from x₃ to x₄):
$$
E_{a,4} = \left| \frac{1.8 - 1.1}{1.8} \right| \times 100\% = \left| \frac{0.7}{1.8} \right| \times 100\% \approx 38.9\%
$$

**Summary Table:**
| iter |   x   |    $E_a$   |
|:----:|:-----:|:------------:|
|  1   | 0.5   |    500%      |
|  2   | 2.2   |   77.3%      |
|  3   | 1.1   |   100%       |
|  4   | 1.8   |   38.9%      |

#### **Explanation:**
- $ E_a $ measures how much the value changes from one iteration to the next, as a percentage of the new value.

---

### Final Answer:

1. **Fixed-point values (x₁, x₂, x₃, x₄):**  
   - $ x_1 \approx 0.5 $
   - $ x_2 \approx 2.2 $
   - $ x_3 \approx 1.1 $
   - $ x_4 \approx 1.8 $

2. **Approximate percent relative errors ($ E_a $) for each step:**  
   - Iter 1: 500%
   - Iter 2: 77.3%
   - Iter 3: 100%
   - Iter 4: 38.9%

**Explanation:**  
The values oscillate and do not rapidly settle to a single value, suggesting the method is not converging quickly (or may be diverging or cycling, depending on further iterations). This matches the graphical behavior, where the steps jump back and forth rather than settle. This is consistent with the fixed-point theorem: convergence depends on the slope of $ g(x) $ at the fixed point. If $ |g'(x^*)| > 1 $, the method will diverge or oscillate.

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