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When using iteration $x_{k+1}=g(x_k),\displaystyle{g(x)=px+\frac{qa}{x^2}+\frac{ra^2}{x^5}}$ in order to find the value of $a^\frac{1}{3}$

What is value of $p,q,r$ that make the sequence converge with highest order as possible?

user76608
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1 Answers1

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Assume $a\neq 0$ because otherwise $g(x)=px$ which we cannot tweak much.

Let $x_\infty$ be the desired limit, so $x_\infty^3=a$. The goal is to maximize the order $m$ in $$\frac{1}{x_\infty}\,g\left(x_\infty (1+\varepsilon)\right) = 1 + \mathrm{O}(\varepsilon^m)$$

Substituting the given Ansatz for $g$ und expanding into power series of $\varepsilon$, we get $$\begin{aligned} \frac{1}{x_\infty}\,g\left(x_\infty (1+\varepsilon)\right) &= p\,(1+\varepsilon) + q\,(1+\varepsilon)^{-2} + r\,(1+\varepsilon)^{-5} \\ &= p\,(1+\varepsilon) + \\ &\quad q \left(1 - 2\varepsilon + 3\varepsilon^2 -4\varepsilon^3 + \mathrm{O}(\varepsilon^4)\right) + \\ &\quad r \left(1 - 5\varepsilon + 15\varepsilon^2 -35\varepsilon^3 + \mathrm{O}(\varepsilon^4)\right) \\ &= \underbrace{(p+q+r)}_1 + \underbrace{(p-2q-5r)}_0\,\varepsilon + \underbrace{(3q+15r)}_0\,\varepsilon^2 + (-4q-35r)\,\varepsilon^3 + \mathrm{O}(\varepsilon^4) \end{aligned}$$ The values under the horizontal braces indicate what we are trying to achieve: We have three parameters $p,q,r$ available for tweaking, therefore we try to set the first three coefficients of the above series to the desired values $1,0,0$. This results in a system of three linear equations in $p,q,r$ which turns out to have the unique solution $$p=q=\frac{5}{9};\quad r=-\frac{1}{9}$$ (If more than one solution had been possible, we would have tried to zero more coefficients.)

With those parameters our relative error estimate becomes $$\frac{1}{x_\infty}\,g\left(x_\infty (1+\varepsilon)\right) = (-4q-35r)\,\varepsilon^3 + \mathrm{O}(\varepsilon^4) = \frac{5}{3}\varepsilon^3 + \mathrm{O}(\varepsilon^4)$$ As demonstrated, any other setting for $p,q,r$ would decrease the order of convergence.

Edit: Typesetting and minor wording corrections.

ccorn
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