Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).

Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).

https://www.gyanodhan.com/video/7A2.%20Computer%20Science/Numerical%20Methods/673.%20Iteration%20method%20%20%28%20with%20it%27s%20convergence%20condition%29.mp4

Iteration Method in Numerical Methods

The Iteration Method is a technique used to find the root of an equation $f(x)=0$. 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 $f(x)=0$ is rewritten in the form:

$$x=g(x)$$

where $g(x)$ is a function derived from $f(x)$.

An initial guess x0x_0x0​ is selected, and the next approximation is obtained using the formula:

xn+1=g(xn)x_{n+1} = g(x_n)xn+1​=g(xn​)

This process continues until the absolute difference between successive approximations is less than a predefined tolerance $ϵ$, i.e.,

∣xn+1−xn∣<ϵ|x_{n+1} – x_n| < \epsilon∣xn+1​−xn​∣<ϵ

2. Convergence Condition

For the iteration method to converge to a root $\alpha$, the function $g(x)$ must satisfy:

∣g′(α)∣<1| g'(\alpha) | < 1∣g′(α)∣<1

where g′(α)g'(\alpha)g′(α) is the derivative of $g(x)$ evaluated at the root $\alpha$.

Explanation of Convergence Condition

• If ∣g′(x)∣<1|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∣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 = 0x3+x−1=0 using iteration method.

Step 1: Rewrite in the form $x=g(x)$

x=1−x3x = 1 – x^3x=1−x3

Step 2: Select an initial guess

Let x0=0.5x_0 = 0.5x0​=0.5.

Step 3: Apply the iteration formula

Using xn+1=1−xn3x_{n+1} = 1 – x_n^3xn+1​=1−xn3​,

x1=1−(0.5)3=1−0.125=0.875x_1 = 1 – (0.5)^3 = 1 – 0.125 = 0.875x1​=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 = 0.3301x2​=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 = 0.9641x3​=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:

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Iteration Method in Numerical Methods

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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.

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Steps to Apply Iteration Method:

1. Rearrange f(x)=0f(x) = 0 into x=g(x)x = g(x)
2. Choose an initial guess x0x_0
3. Compute:

   x1=g(x0),x2=g(x1),…x_1 = g(x_0), \quad x_2 = g(x_1), \quad \dots

4. Stop when:

   ∣xn+1−xn∣<ε|x_{n+1} – x_n| < \varepsilon(where ε\varepsilon is a small tolerance like 0.0001)

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Convergence Condition

For the iteration method to converge to a root α\alpha, the function g(x)g(x) must satisfy:

1. α=g(α)\alpha = g(\alpha) (the fixed point)
2. 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.

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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.

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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

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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	(

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Computer Science/Numerical Methods/ Iteration method ( with it’s convergence condition).

Numerical Methods: Problems and Solutions