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I have struggle on finding this function g(x).

Assume function $f(x) = 5x^3 -20x + 3$ and it is specified to find root in [0, 1]. So I guess, first thing is to find function g(x).

$$g_1(x) = \sqrt[3]{(4x - \frac{3}{5})}$$

$$|g'(x)| = \left| \frac {4} {3\sqrt[3]{(4x-\frac 3 5)^2}}\right| \le \frac 4 3 \gt 1$$

also $f(x) = x(5x^2-20)+3$ and $$g_2(x) = -\frac {3} {5x^2 - 20}$$

but I do not know which one to choose. My teacher said something like

$$|g'(x)| \lt 1 $$

So which one should I choose and why, or $g_1(x)$ and $g_2(x)$ are not good.

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1 Answer 1

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These are not the only choices. In fact, any function $g(x)=k f(x) + x$ would meet the fixed point condition. The most obvious for me is $g_3(x)=\frac{1}{20} ( 5x^3 + 3)$ where it is easy to check the convergence criterium $|g'(x)|<1$.

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  • $\begingroup$ also does $|g'(x)| < 1$ needs to be satisfied only on [0, 1]? that would mean $D = [0, 1], g(D) \subset D$, right? $\endgroup$
    – clzola
    Aug 15, 2015 at 12:15
  • $\begingroup$ @clzola Actually, $|g'(x)| < 1$ on $D = [0,1]$ does not necessarily mean that $g(D) \subset D$. This is a common mistake. $|g'(x)|< 1$ means that there exists some neighborhood of the root $D' = [x_0 - \epsilon, x_0 + \epsilon]$ in which $g(D') \subset D'$. You need to check whether $g$ actually maps $D$ onto itself prior to using fixed point interations. $\endgroup$
    – uranix
    Aug 15, 2015 at 14:01

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