**Iteration Method**

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x_{0} be an initial approximation of the required root α then the first approximation x_{1} is given by x_{1} = pi(x0).

Similarly for second, thrid and so on. approximation

x_{2} = pi(x_{1})

x_{3} = pi(x_{2})

x_{4} = pi(x_{3})

x_{n} = pi(x_{n-1})

**Iteration Method Example:**

Find the real root of the equation x^{3} + x ^{2} = 1 by iteration method.

Solution:

We can rewrite the above equation by

x^{3} + x ^{2} - 1 = 0;

Let f(x) = x^{3} + x ^{2} - 1

f(0) = -1 (positive)

f(1) = 1 (negative)

Hence the root value lie between 0 to 1

x^{3} + x ^{2} - 1 = 0

x^{2} (x + 1) = 1

x^{2} = 1/ (x + 1)

x = 1/ √(x + 1)

pi(x) = 1/ √(x + 1)

Let the initial approximation be x_{0} = 0.5

x_{1} = pi(x_{0}) = 1/√1+ 0.5 = 0.81649

x_{2} = pi(x_{1}) = 1/√1+ 0.81649 = 0.74196

x_{3} = pi(x_{2}) = 1/√1+ 0.74196 = 0.75767

x_{4} = pi(x_{3}) = 1/√1+ 0.75767 = 0.75427

x_{5} = pi(x_{4}) = 1/√1+ 0.75427 = 0.75500

x_{6} = pi(x_{5}) = 1/√1+ 0.75500 = 0.75485

x_{7} = pi(x_{6}) = 1/√1+ 0.75485 = 0.75488

Since the difference between x_{6} and x_{7} are very small, so the root is 0.75488.

**Iteration method Practice problem:**

1. Solve by iteration method 2x - logx - 7 = 0

2. Find the root of the equation x log x = 1.2 by iteration method

3. Compute the real root of 3x - cosx - 1 = 0 by iteration method

4. Find the root of the equation sin x = 1 + x3 between ( -2,-1) to 3 decimal places by Iteration method

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