Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).
Contents
- 1 2. Convergence Condition
- 1.1 Explanation of Convergence Condition
- 1.2 3. Example of Iteration Method
- 1.3 Find the root of x3+x−1=0x^3 + x – 1 = 0x3+x−1=0 using iteration method.
- 1.4 4. Advantages and Disadvantages
- 1.5 Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).
- 1.6 Numerical Methods for Solving Systems of Nonlinear …
- 1.7 MCA-08 / BCA-12: Computer Oriented Numerical Methods
- 1.8 B.Tech 4th Semester MATHEMATICS- …
- 1.9 Numerical Analysis
Iteration Method in Numerical Methods
The Iteration Method is a technique used to find the root of an equation f(x)=0f(x) = 0. It is based on approximating the root through successive iterations, refining the value until it reaches the desired accuracy.
1. General Form of Iteration Method
The given equation f(x)=0f(x) = 0 is rewritten in the form:
x=g(x)x = g(x)
where g(x)g(x) is a function derived from f(x)f(x).
An initial guess x0x_0 is selected, and the next approximation is obtained using the formula:
xn+1=g(xn)x_{n+1} = g(x_n)
This process continues until the absolute difference between successive approximations is less than a predefined tolerance ϵ\epsilon, i.e.,
∣xn+1−xn∣<ϵ|x_{n+1} – x_n| < \epsilon
2. Convergence Condition
For the iteration method to converge to a root α\alpha, the function g(x)g(x) must satisfy:
∣g′(α)∣<1| g'(\alpha) | < 1
where g′(α)g'(\alpha) is the derivative of g(x)g(x) evaluated at the root α\alpha.
Explanation of Convergence Condition
- If ∣g′(x)∣<1|g'(x)| < 1 in the neighborhood of the root, the successive approximations get closer to the root.
- If ∣g′(x)∣>1|g'(x)| > 1, the method diverges, meaning it moves away from the actual root.
3. Example of Iteration Method
Find the root of x3+x−1=0x^3 + x – 1 = 0 using iteration method.
Step 1: Rewrite in the form x=g(x)x = g(x)
x=1−x3x = 1 – x^3
Step 2: Select an initial guess
Let x0=0.5x_0 = 0.5.
Step 3: Apply the iteration formula
Using xn+1=1−xn3x_{n+1} = 1 – x_n^3,
x1=1−(0.5)3=1−0.125=0.875x_1 = 1 – (0.5)^3 = 1 – 0.125 = 0.875 x2=1−(0.875)3=1−0.6699=0.3301x_2 = 1 – (0.875)^3 = 1 – 0.6699 = 0.3301 x3=1−(0.3301)3=1−0.0359=0.9641x_3 = 1 – (0.3301)^3 = 1 – 0.0359 = 0.9641
This process continues until convergence.
4. Advantages and Disadvantages
Advantages:
- Simple and easy to implement.
- Does not require complex calculations like derivatives.
Disadvantages:
- May not always converge.
- Requires a good initial guess.
If you need more examples or explanations, let me know!