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I'm trying to determine a rate of convergence for a non linear function $f(x)=x^5 + 12x^3 -130$ to find its root. Using the fixed point iteration, I am using the second form function $g(x)=(4x^5 + 36x^3 + 130)/(5x^4 + 48x^2)$. If I continue to try values in $g(x)$, eventually I get to the fixed point ~$2.00887$, which is the root of $f(x)$.

What I don't understand is how to determine the rate of convergence to get ~$2.00887$. From the rate of convergence wiki page, I see the formula with $μ$ in it, and know that $q$ is the rate of convergence.

But if you wan't to determine the rate of convergence, how do you know what to set $q$ and $μ$?

If I try the limit, using $x_{k+1} = g(g(2.00887))$ and $x_k = g(2.00887)$ and $q=1$, then $μ$ ends up being ~$1.17$. Does this mean its a linear rate?

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Let $\, r \approx 2.0087496\,$ be the positive real root of $\, f(x) := x^5 + 12x^3 -130 =0\,$ and $$ g(x) := (4x^5+36x^3+130)/(5x^4+48x^2).$$ Then $$g(r+x) \approx 0.1760101\, x^1 + 0.505639\, x^2 + O(x^3).$$ Thus $q=1$ with linear convergence rate $\,\mu \approx 0.1760101.$ In your example, let $$ x_1=2.00887,\quad x_2=g(x_1)\approx 2.00887408,\quad x_3=g(x_2)\approx 2.00887480.$$ Then the quotient of differences from the root is $$ (r-x_2)/(r-x_1) \approx 0.1760076, \quad (r-x_3)/(r-x_2) \approx 0.1760096$$ which are good approximations to $\mu.$

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