Fixed point iteration method converging to infinity

bisectionfixed-point-theoremsnumerical methods

So I am trying to use Fixed Point Iteration Method for this function:

$ f(x)=x^5 -x^4-x-3 $ .

By using bisection method I found out that the root is around $1.64056$.

So I started by doing:

$f(x)=0$
then found out that:

$x=x^5-x^4-3$ .

So here I used Microsoft Excel and it generated very big numbers, which are converging to Inf almost instantly, even if I select my $x_0$ really close to my actual root I need to find. I am adding an Excel screenshot of what I am getting. Is it even possible to use Iteration method for this kind a function? If so, where can I go from here?

enter image description here

Best Answer

So I am trying to use Fixed Point Iteration Method for this function: $$f(x)=x^5 -x^4-x-3$$

To achieve what?

In case you want to find a zero of $f(x)$ by means of an iteration of the form $$x\mapsto g(x)\qquad\text{where}\quad x_0=g(x_0)$$

$x_0$ is a fixed point of $g$, but it's only an attractive fixed-point (assuming $g$ is smooth) when $$|g'(x_0)|<1$$

You chose $g(x) = f(x) + x$ with $x_0\approx 1.64$, $g'(x) = 5x^4-4x^3$ and $$g'(x_0)\approx18.55$$

Thus the fixed-point is clearly repelling.

You can use fixed-point iteration in principle, but as I wrote the absolute value of the derivative at the fixed-point must be less than one$^1$. So you'd have to construct some other function like $$g(x) = \frac{x+3}{x^4} +1$$ (I did not check the derivative condition for this choice, though.$^3$)

Newton-Raphson iteration usually works really well and has quadratic convergence$^2$, in your case: $$x\mapsto x-\frac{f(x)}{f'(x)} = x - \frac{x^5 -x^4-x-3}{5x^4 - 4x^3-1}$$ yields $x_0\approx 1.640585621426312$.

$^1$Equal to one is called "indifferent"; it might converge (slowly) but that case needs further analysis.

$^2$Provided you are "close" enough to the zero and it's not a multiple zero of $f$.

$^3$I did now and it's $-1.42$, so it won't work, either.

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