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How can find the values of $\alpha$ and $\beta$, such that $x_{n+1}=\alpha x_n\left(3-\dfrac{x_n^2}{a}\right)+\beta x_n\left(1+\dfrac{a}{x_n^2}\right)$ has $3^{\text{rd}}$ order convergence to $\sqrt a$ ?

  1. $\alpha=\dfrac{3}{8}$ and $\beta=\dfrac{1}{8}$
  2. $\alpha=\dfrac{1}{8}$ and $\beta=\dfrac{3}{8}$
  3. $\alpha=\dfrac{2}{8}$ and $\beta=\dfrac{2}{8}$
  4. $\alpha=\dfrac{1}{4}$ and $\beta=\dfrac{3}{4}$
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    What did you try? Context? – 2017-02-21
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    $x_{n+1}=f(x_n)$ is converges if absolute value of f'(x)<1 – 2017-02-21
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    But you need to find $3^{rd}$ order of convergence. Hence you have to prove that $\frac{|x_{k+1}-\sqrt{a}|}{(x_k-\sqrt{a})^3}$ converges – 2017-02-21
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    which of the following is possible (1) $\alpha=\frac{3}{8}$ and $\beta=\frac{1}{8}$ (2) $\alpha=\frac{1}{8}$ and $\beta=\frac{3}{8}$ (3) $\alpha=\frac{2}{8}$ and $\beta=\frac{2}{8}$ (4) $\alpha=\frac{1}{8}$ and $\beta=\frac{3}{8}$ – 2017-02-21
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    Why $(2)$ and $(4)$ are the same? – 2017-02-21

1 Answers 1

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Set $x_{n+1}=f(x_n)=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a^2}{x_n^2})$ and denote $z=\sqrt{a}$. Replace $x_n$ by $x$.

Using Taylor's development we find:

$$f(x)=f(z)+f'(z)(x-z)+\frac{f''(z)}{2}(x-z)^2+\frac{f'''(z)}{6}(x-z)^3+\frac{f''''(\xi)}{24}(x-z)^4$$

for some $\xi\in(x,z)$. Since $f(z)=z$ and dividing the previous equation by $|x-z|^3$ we have:

$$\frac{f(x)-f(z)}{|x-z|^3}=\frac{f'(z)}{|x-z|^2}+\frac{f''(z)}{2|x-z|}+\frac{f'''(z)}{6}+\frac{f''''(\xi)}{24}|x-z|$$

So in order to converge to the $3^{rd}$ order we have to impose both $f'(z)=0$ and $f''(z)=0$. This give raise to the following system of equations:

$$ \begin{align} \beta(1-a)&=0\\ -3\alpha+\beta&=0 \end{align} $$

and hence

$$ \begin{align} \beta(1-a)&=0\\ \beta&=3\alpha& \end{align} $$

Looking at the proposed solutions we have to choose $\color{red}{\boxed{\color{black}{(2)}}}$

Remark: solution $(4)$ seems to be a duplicate of $(2)$.

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    Thank you so much sir, nice explanation – 2017-02-22