The iterates xn+1 = 2 − (1 + c)xn + cx3
n will converge to α = 1 for some values of constant c (provided that
x0 is sufficiently close to α). Find the values of c for which convergence occurs? For what values of c, if any,
convergence is quadratic.

The given iteration formula is:

\[
x_{n+1} = 2 - (1 + c)x_n + c x_n^3
\]

We are tasked with determining the values of \( c \) for which the iterates converge to \( \alpha = 1 \) and to check for quadratic convergence.

### Step 1: Analyzing Convergence to \( \alpha = 1 \)

The iterates \( x_n \) are said to converge to \( \alpha = 1 \). For convergence, the fixed point condition at \( \alpha = 1 \) must be satisfied. Let's substitute \( \alpha = 1 \) into the iteration formula and check if it holds:

\[
x_{n+1} = 2 - (1 + c) \cdot 1 + c \cdot 1^3
\]

Simplifying:

\[
x_{n+1} = 2 - (1 + c) + c = 2 - 1 = 1
\]

Thus, \( \alpha = 1 \) is indeed a fixed point, i.e., if \( x_n = 1 \), then \( x_{n+1} = 1 \).

### Step 2: Linearization Near the Fixed Point

To analyze convergence, we linearize the iteration near \( \alpha = 1 \). Let \( x_n = 1 + \epsilon_n \), where \( \epsilon_n \) is a small perturbation around the fixed point. Substituting into the iteration formula:

\[
x_{n+1} = 2 - (1 + c)(1 + \epsilon_n) + c(1 + \epsilon_n)^3
\]

Expand the terms up to first order in \( \epsilon_n \):

\[
x_{n+1} \approx 2 - (1 + c)(1 + \epsilon_n) + c(1 + 3\epsilon_n) = 2 - (1 + c)(1 + \epsilon_n) + c(1 + 3\epsilon_n)
\]

Simplifying:

\[
x_{n+1} \approx 2 - (1 + c) - (1 + c)\epsilon_n + c + 3c\epsilon_n
\]

\[
x_{n+1} \approx 1 - (1 + c)\epsilon_n + 3c\epsilon_n
\]

\[
x_{n+1} \approx 1 + (3c - (1 + c))\epsilon_n
\]

\[
x_{n+1} \approx 1 + (2c - 1)\epsilon_n
\]

Thus, the linearized iteration becomes:

\[
\epsilon_{n+1} \approx (2c - 1) \epsilon_n
\]

### Step 3: Condition for Convergence

For convergence, the perturbation \( \epsilon_n \) must decay, meaning \( |2c - 1| < 1 \). Solving this inequality:

\[
-1 < 2c - 1 < 1
\]

\[
0 < 2c < 2
\]

\[
0 < c < 1
\]

Therefore, the iterates converge to \( \alpha = 1 \) if \( c \) is in the interval \( (0, 1) \).

### Step 4: Checking for Quadratic Convergence

Quadratic convergence occurs if the first-order term in \( \epsilon_n \) vanishes. This happens when the coefficient of \( \epsilon_n \) in the linearized equation is zero:

\[
2c - 1 = 0
\]

Solving for \( c \):

\[
c = \frac{1}{2}
\]

Thus, for \( c = \frac{1}{2} \), the convergence is quadratic.

### Final Answer:

- **The iterates converge to \( \alpha = 1 \) for \( 0 < c < 1 \).**
- **The convergence is quadratic when \( c = \frac{1}{2} \).**

Would you like more details on any part of this explanation?

### Further Questions:
1. How can we determine the rate of convergence in general?
2. What happens if \( c \) is outside the interval \( (0, 1) \)?
3. How do we formally define quadratic convergence in numerical methods?
4. Can this iteration be generalized for different values of \( \alpha \)?
5. What is the behavior of the iteration formula for large values of \( n \)?

### Tip:
Linear convergence occurs when the error is reduced proportionally at each step, while quadratic convergence is faster, reducing the error roughly as the square of the previous error.

The iterates xn+1 = 2 − (1 + c)xn + cx3n will converge to α = 1 for some values of constant c (provided that x0 is sufficiently close to α). Find the values of c for which convergence occurs? For what values of c, if any, convergence is quadratic.

Explore the convergence of the iteration xn+1 = 2 − (1 + c)xn + cx^3n to the fixed point α = 1. Learn about the values of c for which convergence occurs, as well as the conditions for quadratic convergence. Suitable for advanced high school or undergraduate students.

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