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Suppose that we have the iteration method $x_{k+1}=f(x_{k})$ and $f(x)=Ax+Bx^{2}+Cx^{3}$.For given $a>0$ , determine $A$, $B$, $C$ in order to have local convergence to $\dfrac{1}{a}$ with convergence rate $p=3$.

So I suppose that we have to calculate $$\lim_{n\rightarrow \infty} \frac{\left | x_{k+1}-\frac{1}{a} \right |}{\left | x_{k}-\frac{1}{a} \right |^{3}}=\lim_{n\rightarrow \infty}\frac{\left | Ax_{k}+Bx_{k}^{2}+Cx_{k}^{3}-\frac{1}{a} \right |}{\left |x_{k}-\frac{1}{a} \right |}=\lambda$$

and determine $A$, $B$, $C$ in such a way that $\lambda$ is finite , but I don't truely see how to do that.

Ng Chung Tak
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Jonathan1234
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1 Answers1

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You need $$ af(x)-1 = c(ax-1)^3=ca^3x^3-3ca^2x^2+3cax-c $$ Try to find out why $c$ has to be a constant if $f$ is to be a cubic polynomial.

As $f$ has no constant term, you immediately find $c=1$ and thus $$ f(x)=a^2x^3-3ax^2+3x. $$

Lutz Lehmann
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  • But how we found $af(x)-1$ ?? We used taylor series ?? Because the only definition that I know for convergence is the one that I wrote in the beginning. – Jonathan1234 Jan 02 '18 at 18:13
  • You need a differentiable function $F$ with $F(1/a)=0$, $F'(1/a)\ne 0$. Here I chose $F(x)=ax-1$. Any other variant works too, but it is easier if you stay in the polynomial realm. Then the condition of cubic converge is equivalent to $F(f(x))=O(F(x)^3)$. – Lutz Lehmann Jan 02 '18 at 20:39