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I'm given a sequence $x_n \rightarrow \alpha$ of a nonlinear solver such that $$\lim_{n\rightarrow\infty}\frac{x_{n+1}-\alpha}{x_n-\alpha}=c$$ converges linearly (i.e. $c\in(0,1)$).

Now, I need to determine the asymptotic rate of convergence $p^*$ and constant $c^*$ of new iterations, determined by the sequence

$$x_n^*=\frac{x_n+x_{n-1}}{2}$$

I've attempted to solve this by using the definition of linear convergence to show that as

$$\lim_{n\rightarrow\infty}\frac{\varepsilon_{n+1}}{\varepsilon_{n}} =c$$

for $\varepsilon_i \geq |x_i - \alpha|$, the asymptotic error constant of $x_n^*$

$$\lim_{n\rightarrow\infty}\frac{x_{n+1}^*-\alpha}{(x_n^* -\alpha)^{p^*}}=c^*$$

is bounded by

$$\lim_{n\rightarrow\infty}\frac{\frac{\varepsilon_{n+1}+ \varepsilon_n}{2}}{(\frac{\varepsilon_{n}+ \varepsilon_{n-1}}{2})^{p^*}}$$

which I found by substituting the values of $x_n^*$ using the variable definition.

The first problem is that I'm not sure if it's correct to simplify further by substituting the initial constant of convergence $c$ to get a result in terms of $\varepsilon_n$. If I do, I would get something like this:

$$\lim_{n\rightarrow\infty}\frac{(c+1)/2}{(\frac{c+1}{2c})^{p^*}\varepsilon_n^{p^* -1}}$$

The second problem is that I'm not sure this answers the question, which I presume is to determine $p^*$ and $c^*$ in terms of the original solver (i.e. they should not include one another).

MrT
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  • Why do you expect that $x^_n$ converges faster than the original sequence? Where does $p^$ originate? – Lutz Lehmann May 04 '15 at 17:31
  • Sorry, I should have updated my question. I later came to the conclusion that in order for $p^$ to represent a converging series, the order must be 1, since otherwise the denominator would become 0 as $n\rightarrow\infty$. In this case, we find that $c^ = c$. So there is no difference between the two. – MrT May 04 '15 at 17:55

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