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Given $x=\sqrt {a}$ is a fixed point of the function $g(x)= \frac{x^3+3xa}{3x^2+a}$. Determine the order of convergence and asymptotic error constant of sequence $P_n =g(P_{n-1})$ towards $x=\sqrt{a}$.

I know the definition of order of convergence but here i have no clue to proceed.

Arthur
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Kavita
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  • Taylor expand $g(x)$ about $x=\sqrt{a}$, retaining the second nonzero term (the first will be of course $g(\sqrt{a})=\sqrt{a})$). – Ian Nov 28 '16 at 15:09
  • Still i m not getting the answer, a bit confusing – Kavita Nov 28 '16 at 15:12
  • What did you get from doing the Taylor expansion? It should look like $g(x)=\sqrt{a}+C(x-\sqrt{a})^k$ plus higher order terms, where $k$ is a positive integer and $C=g^{(k)}(\sqrt{a})/k!$. Then $k$ and $C$ are the answers to your question. – Ian Nov 28 '16 at 15:12
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    You can very easily show that $g(x)-\sqrt{a}=\frac{(x-\sqrt{a})^3}{3x^2+a}$. – Lutz Lehmann Dec 02 '16 at 04:34

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