# Convergence with Fixed Point Equations

The goal is to solve the root finding problem using fixed point equations.

I am somewhat stuck here. I have derived several forms of the following equation: $$f(x) = x^3 - 2x + 1$$ The forms I was meant to derive include:

• $g(x) = 2/x - 1/x^2$ which I was able to simplify to $(2x - 1)/x^2$

• $g(x) = \sqrt{2 - 1/x}$

• $g(x) = -(1-2x)^{1/3}$

The initial point $p_o$ is $0.5$. However, I cannot seem to get the equations to iterate to show convergence as no matter what I have tried for manipulations has resulted in the same equation. I have multiplied by conjugates and attempted to remove denominators. Trying the simplify function in wolfram alpha was of no help. I am supposed to derive the fixed point equations above and then determine if they converge. I am not a terrific mathematician but thought I could at least manipulate the above equations. Am I missing or not understanding something? How can I manipulate the equations to find roots or divergence instead of $0$ and then an undefined result?

My first attempt was on a simpler problem and worked easily so I think I understand fixed point iteration to a certain extent at least.

Any help is appreciated. Thanks.

• Are you trying to see if the the sequence $(x_n)_{n\geq0}$ defined by $x_0=1/2$, $x_{n+1}=f(x_n)$ for $n\geq 0$ is convergent? – Robert Z Sep 7 '16 at 7:32
• @Robert Z I am trying to find the roots using fixed point equations where x+1 = g(x) – Andrew Scott Evans Sep 7 '16 at 7:43
• The roots of $x^3 - 2x + 1=0$? – Robert Z Sep 7 '16 at 7:45
• @Robert Z Yes. I know the roots to be x = 1, x = .618, and x= -1.1618 but, starting at .5 as x0, must find if each g(x) converges to a root in an iterative fashion. However, g(.5) = 0 for each equation and g(0) then is undefined. Is there a way to manipulate them to make each one work without resorting to the already solved 1/2*x^3 + 1/2 – Andrew Scott Evans Sep 7 '16 at 7:49
• @Robert Z I need to find if they converge either by finding a root or showing that they cannot find any root as the roots are not given. – Andrew Scott Evans Sep 7 '16 at 7:51

Perhaps this comment is not very pertinent but, as I understand, the fixed point theorem is relevant in the uniqueness. But your function $f(x)=x^3-2x+1$ admits three fixed points. You have (approximately) $$f(x)=x$$ for the three values $$x\approx -1.879\\x\approx 0.347\\x\approx 1.532$$