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I would appreciate help interpreting the suggested solution for the task below, I cannot see how they get that $\varphi(x)=x-\frac{1}{2}u(x)-\frac{1}{2}[u(x)/u^\prime(x)]$:

Question:

$f(x)$ is a four times continuously differentiable and have a simple zero $\xi$. Successive approximations $x_n,n=1,2,...\xi$ are computed from $x_{n+1}=(x´_{n+1}+x´´_{n+1})/2$, where $$x´_{n+1}=x_n-\frac{f(x_n)}{f^\prime(x_u)},$$ $$x´´_{n+1}=x_n-\frac{u(x_n)}{u^\prime(x_u)}, \text{ } u(x)=\frac{f(x)}{f^\prime(x)}.$$

Prove that if the sequence $\{x_n\}$ converges to $\xi$ then the convergence is cubic.

Solution:

enter image description here

And the last line shows that the line is at least cubic.

I've been looking at the solution for a while now and feel that I understand most things but one thing I do not understand is how they can get that $\varphi(x)=x-\frac{1}{2}u(x)-\frac{1}{2}[u(x)/u^\prime(x)]$?

I first thought that they simply put in the expressions for $x´$ and $x´´$ in $x_{n+1}=(x´_{n+1}+x´´_{n+1})/2$ but in that case, what happens with the first $x_n$ in both $x´$ and $x´´$?

kabin
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1 Answers1

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It's still there. They just added the two $x_n$s together and then divided by 2. So you get \begin{equation*} x_{n+1} = (x_{n+1}' + x_{n+1}'')/2 = \frac{1}{2}\bigg(x_n-\frac{f(x_n)}{f'(x_n)} + x_n - \frac{u(x_n)}{u'(x_n)}\bigg) = \frac{1}{2}\bigg(2x_n-\frac{f(x_n)}{f'(x_n)} - \frac{u(x_n)}{u'(x_n)}\bigg) = \frac{1}{2}\bigg(2x_n-u(x_n) - \frac{u(x_n)}{u'(x_n)}\bigg) \end{equation*} Then as they wrote, we have $x_{n+1} = \phi(x_n)$. Then you just set $ x = x_n = x_{n+1}$

jbenoit
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