Fixed point for a function. Numerical Analysis.

Question

how might g defined such that the root of $f(x)=x^3-3x-1$ is a fixed point of $g$, for $x$ in the closed interval between $-\frac{1}{2}$ and $0$. Find two distinct functions $g_1$ and $g_2$ and investigate the rate of convergence for each case.

I can find three different such functions for $g$, do I just pick one or do I have an extra one by mistake? My functions: $g_1(x)=(x^3-1)/3$, $g_2(x)=(1+3x)^{1/3}$ and $g_3(x)=1/(x^2-3)$. (also because I'm new to this website, can someone please link me a guide to use latex so that my question will look less confusing hopefully!)

Answer 1

All three of these functions are good ones to have a fixed point where you want. Yes, you can pick any two. The root of interest is about $-0.3473$. It might be nice to choose one stable and one unstable fixed point, but the question does not require that.

Answer 2

$g_1$ converges linearly (in the nomenclature used in Numerical recipes) with a constant of about $0.1206$ (that is, if $x_0$ is the fixed point, then for small $\delta$ $$ g_1 : x_0 + \delta \rightarrow x_0 + 0.1206 \,\delta $$ $g_3$ does a bit better but still converges linearly: $$ g_3 : x_0 + \delta \rightarrow x_0 + 0.0838 \,\delta $$ There is a linear combination of $g_1$ and $g_3$ that converges super-linearly: $$ g_4(x) = 3.274316\ldots g_3(x) - 2.274316\ldots g_1(x) \\ g_4 : x_0 + \delta \rightarrow x_0 + 0.3288 \,\delta^2 $$

Answer 3

Hint: For a $g$ with quadratic convergence, try Newton's method for $f$.