The rate of convergence represents how quickly a sequence approaches its limit.
$$ \lim _{n \rightarrow \infty} \frac{\left|x_{n+1}-x^{*}\right|}{\left|x_{n}-x^{*}\right|^{q}}=\mu $$
Here, the rate of convergence is μ
Looking at convergence speed for iterative methods, we know that for a sequence to converge Q-linearly, we see:
$$ \lim _{k \rightarrow \infty} \frac{\left|x_{k+1}-L\right|}{\left|x_{k}-L\right|}=\mu $$
where the sequence x converges to the number L.
Can this (rate of convergence) be computed by taking the slope (or 2nd derivative) of a plotted graph that shows the convergence of the sequence (rather than calculating it analytically)?
The question is about a sequence where $$ \lim_{n\to\infty} \frac{|x_{n+1}-L|}{|x_{n}-L|^q}=\mu. \tag{1} $$ Define the associated sequence $$ y_n := \log|x_{n}-L|. \tag{2} $$ Now equation $(1)$ can be written as $$ \lim_{n\to\infty} (y_{n+1} - q\,y_n) =\log\mu . \tag{3} $$ In the case where $\,q=1\,$ the condition is $$ \lim_{n\to\infty} (y_{n+1} - \,y_n) =\log\mu . \tag{4} $$ When $\,n\,$ is big this is $$ y_{n+1} \approx y_n + \log\mu. \tag{5} $$ The question asks
If the sequence $\,y_n\,$ is plotted on a graph, then the slope of the line that best approximates the sequence is a good approximation to $\,\log\mu\,$ which gets better as $\,n\,$ gets bigger.
If $\,q>1\,$ then use $\,\log\,y_n\,$ instead.