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The fixed point method $g(x)= x - \frac{f(x)}{2f ^\prime (x)}$ (can also be written as $x_{k+1} = x_k - \frac{f(x_k)}{2f ^\prime (x_k)}$) is similar to Newton's method except for the 1/2 coefficient. My question is how do I show the order of convergence for this method using Taylor series?

EDIT

I believe the goal is to get it into the form $|x_{k+1} - a| = C|x_{k} - a|^n$ and that $n$ is the order of convergence.

I tried getting it into the form $g(y) = g(x) + g^\prime(x)(y-x) + \frac{g ^{\prime\prime}(x)}{2!}(y-x)^2 + ...$

Since $g^\prime (x) = \frac{1}{2} + \frac{1}{2} \cdot \frac{f ^{\prime \prime} (x) f(x)}{[f^\prime (x)]^2}$

Substituting in I get $g(x_k) = g(a) + \frac{1}{2}(x_k -a) + \frac{g^{\prime \prime}(a)}{2}(x_k-a) $

But I'm not sure what to do from here

user501880
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1 Answers1

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Your last formula should be $$ g(x_k) = g(a) + \frac{1}{2}(x_k -a) + \frac{g''(\tilde a)}{2}(x_k-a)^2 $$ Now $g(a)=a$ and thus $$ |x_{k+1}-a|= \frac{1}{2}|x_k -a|+O(|x_k-a|^2) $$ which is indicative of linear convergence.

Choosing a $C\in(\frac12,1)$ and $N$ large enough, the above equation can be weakened to $$ |x_{k+1}-a|\le C\,|x_k -a| $$ for all $k\ge N$ resp. $x_k$ close enough to $a$.

Lutz Lehmann
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