I have an equation $f(x)= x^4 +2x^3 -5x^2 -7x-5=0 $ and I want to solve it (numerically) at [1.5,3] using the fixed point iteration-method, so I want to find a function g(x) such that $f(x)=0$ is equivalent to $g(x)=x$. However I've tried many different $g(x)$'s but can't seem to find one that converges to the solution (which is about 2.19).
First of all, anyone has any idea which $g(x)$ I could use?
Second is there a trick to finding it?
Thanks in advance 🙂
[Math] Fixed point iteration-finding $g(x)$
fixed-point iterationfixed-point-theoremsnumerical methods
Best Answer
To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| \lt 1$ 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 $$x^4 +2x^3 -5x^2 -7x-5=0\\x^4=-2x^3+5x^2+7x+5\\ x=\sqrt[4]{-2x^3+5x^2+7x+5}$$ 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 $$x^4 +2x^3 -5x^2 -7x-5=0\\x^4=-2x^3+5x^2+7x+5\\ x=(-2x^3+5x^2+7x+5)/x^3$$ 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.