Consider finding a numerical solution to

 

 

f(x)=x3+2x2−4=0.
 
 

 

ising the iterative scheme

 

 

xk+1=g(xk)=4xk+2−−−−−−√,k=0,1,…
 
 

(a) Using an initial guess  x0=1
 , calculate  x2
  to two decimal places.

 

x2=
  
1.13
 

 

(b) Theory guarantees that the fixed-point iteration converges to a solution for every initial guess  x0
  greater than what minimum value (accurate to two decimal places)?

 

x0≥
  
-2
 

Let's solve the given problem step by step.

### (a) Calculate \( x_2 \) using the iterative scheme:

We are given the iterative scheme:

\[
x_{k+1} = g(x_k) = \sqrt{4x_k + 2}, \quad k = 0,1,\dots
\]

With the initial guess \( x_0 = 1 \), we want to calculate \( x_2 \).

1. **Calculate \( x_1 \):**
   \[
   x_1 = \sqrt{4x_0 + 2} = \sqrt{4(1) + 2} = \sqrt{4 + 2} = \sqrt{6}
   \]
   \[
   x_1 \approx 2.45
   \]

2. **Calculate \( x_2 \):**
   Using \( x_1 \approx 2.45 \),
   \[
   x_2 = \sqrt{4x_1 + 2} = \sqrt{4(2.45) + 2} = \sqrt{9.8 + 2} = \sqrt{11.8}
   \]
   \[
   x_2 \approx 3.43
   \]

Thus, \( x_2 \approx 3.43 \).

### (b) Minimum initial guess \( x_0 \):

We need to find the minimum value of \( x_0 \) such that the fixed-point iteration converges.

The function \( g(x) = \sqrt{4x + 2} \) will converge if the initial guess is greater than or equal to a certain value. We set the inequality to ensure that \( x_0 \geq -2 \), because when \( x = -2 \):

\[
g(x) = \sqrt{4(-2) + 2} = \sqrt{-8 + 2} = \sqrt{-6}
\]

At \( x = -2 \), the function becomes imaginary, so the iteration will break down for values less than \( -2 \). Therefore, for real solutions, the iteration will work for \( x_0 \geq -2 \).

Thus, the minimum value of \( x_0 \) is:
\[
x_0 \geq -2
\]

### Summary:
- \( x_2 \approx 3.43 \)
- \( x_0 \geq -2 \)

Would you like more details, or do you have any other questions? 

Here are five related questions:
1. What happens to the iteration process for values of \( x_0 \) less than -2?
2. How does the convergence behavior change for initial guesses much larger than -2?
3. Can we use other iterative methods for the same equation?
4. What is the general criterion for the convergence of fixed-point iterations?
5. How would the solution change if the function \( f(x) \) were different?

**Tip:** Fixed-point iteration converges faster when the derivative of \( g(x) \) is small near the root.

Consider finding a numerical solution to f(x) = x^3 + 2x^2 − 4 = 0 using the iterative scheme x_(k+1) = g(x_k) = sqrt(4x_k + 2), k = 0,1,... (a) Using an initial guess x_0 = 1, calculate x_2 to two decimal places. (b) Theory guarantees that the fixed-point iteration converges to a solution for every initial guess x_0 greater than what minimum value (accurate to two decimal places)?

Learn how to find a numerical solution to the equation f(x) = x^3 + 2x^2 − 4 using the fixed-point iteration scheme x_(k+1) = sqrt(4x_k + 2). This detailed step-by-step guide demonstrates how to calculate x_2 with an initial guess of x_0 = 1 and discusses the convergence conditions for fixed-point iteration. Suitable for undergraduate students.

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