Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).
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
- 2 📘 Iteration Method in Numerical Methods
- 3 🧠 Summary Table:
- 4 📥 Want More?
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!
Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).
Numerical Methods for Solving Systems of Nonlinear …
MCA-08 / BCA-12: Computer Oriented Numerical Methods
B.Tech 4th Semester MATHEMATICS- …
Numerical Analysis
Here’s a complete and easy-to-understand explanation of the Iteration Method in Numerical Methods (especially useful for Computer Science or Engineering students), along with its convergence condition:
📘 Iteration Method in Numerical Methods
✅ What is the Iteration Method?
The Iteration Method (also called the Successive Approximation Method or Fixed Point Iteration Method) is used to find the root of a non-linear equation:
f(x)=0f(x) = 0
We rewrite this into the form:
x=g(x)x = g(x)
Then we apply the iteration:
xn+1=g(xn)x_{n+1} = g(x_n)
We continue this process until the difference ∣xn+1−xn∣|x_{n+1} – x_n| is less than a desired tolerance.
🔁 Steps to Apply Iteration Method:
- Rearrange f(x)=0f(x) = 0 into x=g(x)x = g(x)
- Choose an initial guess x0x_0
- Compute:
x1=g(x0),x2=g(x1),…x_1 = g(x_0), \quad x_2 = g(x_1), \quad \dots
- Stop when:
∣xn+1−xn∣<ε|x_{n+1} – x_n| < \varepsilon(where ε\varepsilon is a small tolerance like 0.0001)
📌 Convergence Condition
For the iteration method to converge to a root α\alpha, the function g(x)g(x) must satisfy:
- α=g(α)\alpha = g(\alpha) (the fixed point)
- Convergence condition:
∣g′(α)∣<1|g'(\alpha)| < 1
This means the derivative of g(x)g(x) near the root should be less than 1 in absolute value.
If ∣g′(x)∣>1|g'(x)| > 1, the method diverges.
📙 Example Problem:
Find the root of the equation:
x3+x−1=0x^3 + x – 1 = 0
✅ Step 1: Rearrange to the form x=g(x)x = g(x).
Let’s take:
x=11+x2x = \frac{1}{1 + x^2}
✅ Step 2: Let x0=0.5x_0 = 0.5
✅ Step 3: Apply iterations:
x1=11+(0.5)2=0.8x_1 = \frac{1}{1 + (0.5)^2} = 0.8 x2=11+(0.8)2≈0.6098x_2 = \frac{1}{1 + (0.8)^2} ≈ 0.6098 x3=11+(0.6098)2≈0.728x_3 = \frac{1}{1 + (0.6098)^2} ≈ 0.728
… and continue until convergence.
✅ Step 4: Check ∣xn+1−xn∣<0.001|x_{n+1} – x_n| < 0.001 for stopping.
📈 Advantages of Iteration Method:
- Simple to understand and implement
- Needs only one initial guess
⚠️ Limitations:
- Convergence is not guaranteed unless condition ∣g′(x)∣<1|g'(x)| < 1 holds
- May converge slowly
🧠 Summary Table:
Term | Description |
---|---|
Equation Form | x=g(x)x = g(x) |
Iteration Rule | xn+1=g(xn)x_{n+1} = g(x_n) |
Convergence | If ( |
Divergence | If ( |
Stopping Rule | ( |
📥 Want More?
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- A PDF handout for this topic?
- Practice problems with solutions?
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