(a) statement of fixed point theorem lein- fixed point iteration fixed point iteration x = g(x) method.

(a) STATEMENT OF FIXED POINT THEOREM lein- FIXED POINT ITERATION x = g(x) METHOD.

1.  Fixed point iteration : x = g(x) method.

The method known is fixed-point iteration (we also call it the x = g(x) method) is a very useful method to get a root of f (x) = 0.

Suppose, we want the approximate roots of the equation f (x) 0, we rearrange f(x) into an equivalent form x = g(x) ….. (1)

Assume, xo to be the starting approximate value to the actual root r (i.e.,) f (r)

Setting x = x₀ in the right hand side of (1), we get first approximation

x₁ = 8 (x₀)

Again setting x = x1 on the R.H.S. of (1), we get successive approximations

x₂  = g (x₁)

x₃ = 8 (x₂)

… … ...

… … ...

x_n = g (x_n – 1)

The sequence of approximate roots x1, x2, ... xn, if it converges to r is taken as the root of the equation f (x) = 0. Whenever, we have x, = g(x), r is said to be a fixed point for the function g.

Note: The convergence of the sequence is not guaranteed always unless the choice of x₀ is properly chosen.

2. The condition for the convergence of the method

Theorem: Let f (x) = 0 be the given equation whose actual root is r. The equation f(x) = 0 can be written as x = g (x) . Let I be the interval containing the root x = r. If g'(x) | < 1 for all x in I, then the sequence of approximations x₀, x₁, x₂, … x_n Will converge to r , if the initial starting value x₀ is chosen in I.

Note 1: Since |x_n – r | ≤ K |x_n-1 - r | where K is a constant the convergence is linear and the convergence is of order 1.

 Note 2: The sufficient condition for the convergence is g' (x) < 1for all x in I (interval)

1. Find a real root of the equation x³ + x² – 1 = 0 by iteration method. [A.U M/J 2012] [A.U N/D 2019 (R-17)]

Solution :

Note : (1) Fix 4 decimal places in your calculator.

(2) If decimal places is not given, we choose 4 decimal places in general.

Let f (x) = x³ + x² - 1

f(0) = 0 + 0-1 = -1 = -ve

f(1) = 1+1-1 = 1 = +ve

a real root lies between 0 and 1.

2. Find the smallest positive root of 3x= √1 + sin x correct to three decimal places by iterative method. [A.U N/D 2010] [A.U N/D 2020 (R-17)] [A.U A/M 2021 (R-17)]

Solution :

Note :

(1) Fix 3 decimal places in your calculator.

(2) Change degree mode to radian mode in your calculator.

3. Find a real root of the equation cos x 3x-1 correct to 5 decimal places by fixed point iteration method. [A.U M/J 2013]

Solution :

Note :

(1) Fix 5 decimal places in your calculator.

(2) Change degree mode to radian mode in your calculator.

4. Solve, e^x - 3x = 0 by the method of fixed point iteration. [AU N/D 2011] [AU M/J 2012]

Solution :

Note : Fix 4 decimal places in your calculator.

5. Find the negative root of the equation x³ -2x + 5 = 0.

Solution:

Note : Fix 4 decimal places in your calculator.

Given: x³- 2x + 5 = 0

Let K (x) = x³-2x+5

To find the negative root is nothing but to find the positive root of

6. Use the method of fixed point iteration to solve the equation nera. 3x - log_{10 x} = 6.

Solution : Note: Fix 4 decimal places in your calculator.

Given: 3x-log_10x = 6