CAS PROJECT. Fixed-Point Iteration. (a) Existence. Prove that if g is continuous in a

Chapter 19, Problem 19.1.28

(choose chapter or problem)

Fixed-Point Iteration.

(a) Existence. Prove that if g is continuous in a closed interval I and its range lies in I, then the equation x = g(x) has at least one solution in I. Illustrate that it may have more than one solution in I.

(b) Convergence. Let \(f(x)=x^{3}+2 x^{2}-3 x-4=0\). Write this as x = g(x), for g choosing (1) \(\left(x^{3}-f\right)^{1 / 3}\), (2) \(\left(x^{2}-\frac{1}{2} f\right)^{1 / 2}\), (3) \(x+\frac{1}{3} f\), (4) \(x\left(1+\frac{1}{4} f\right)\), (5) \(\left(x^{3}-f\right) / x^{2}\), (6) \(\left(2 x^{2}-f\right) / 2 x\), (7) \(x-f / f^{\prime}\) and in each case \(x_{0}=1.5\). Find out about convergence and divergence and the number of steps to reach exact 6S - values of a root.

Text Transcription:

f(x) = x^3 + 2x^2 - 3x - 4 = 0

(x^3 - f)^1/3

(x^2 - 1 / 2 f)^1/2

x + 1 / 3 f

x (1 + 1 / 4 f)

(x^3 - f) / x^2

(2x^2 - f)/2x,

x - f/f ‘

x_0 =1.5