Question

find point iteration method written in the firm​

Answer 1

Answer:

To get the solution to converge, besides g(x)=x you need |g′(x)|<1 at the root. The distance from the root is multiplied by about |g′(x)| every iteration, so if that is less than 1 it will converge. Powers change quickly and roots slowly, so a natural try is to write

x4+2x3−5x2−7x−5=0x4=−2x3+5x2+7x+5x=−2x3+5x2+7x+5−−−−−−−−−−−−−−−−−√4

It converges nicely starting at 0,1 and 3.5 but fails starting from 4 because the piece under the root goes negative. Another try, which I have not tested, would be to write

x4+2x3−5x2−7x−5=0x4=−2x3+5x2+7x+5x=(−2x3+5x2+7x+5)/x3

It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed g(x) at it.

Step-by-step explanation:

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Answer 2

Answer:

Fixed Point Iteration Method : In this method, we first rewrite the equation (1) in the form

x = g(x) (2)

in such a way that any solution of the equation (2), which is a fixed point of g, is a solution of

equation (1). Then consider the following algorithm.

Algorithm 1: Start from any point x0 and consider the recursive process

xn+1 = g(xn), n = 0, 1, 2, ...