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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!)

Mark Fischler
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UnsinkableSam
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    Enclose equations in dollar signs. Use x_1 for the subscript in $x_1$. Use your exponents just the way you did, but if they are more than one character, enclose them in curly brackets. If you edit the question in a minute, you will see how this looks. – Mark Fischler Mar 18 '19 at 20:32
  • great! thank you so much! – UnsinkableSam Mar 18 '19 at 22:22

3 Answers3

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$g_1$ 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: $$ g_4(x) = 3.274316\ldots g_3(x) - 2.274316\ldots g_1(x) \\ g_4 : x_0 + \delta \rightarrow x_0 + 0.3288 \,\delta^2 $$

Mark Fischler
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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.

Ross Millikan
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  • I would be reluctant to use $g_2$ as an example to discuss convergence rate since it does not converge to the desired fixed point for any open region of starting values of $x$. – Mark Fischler Mar 18 '19 at 21:19
  • @MarkFischler: I think it is a good example just because it doesn't converge. OP is being asked to compare them He could do the same analysis as you do and show that the distance from the root is multiplied by something greater than $1$. But using the two examples you pick answers the question as well. – Ross Millikan Mar 19 '19 at 00:22
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Hint: For a $g$ with quadratic convergence, try Newton's method for $f$.

lhf
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