what is the meaning of asymptotic error constant

Question

I got the definition of asymptotic error constant: $$A=\lim_{n\to\infty} \frac{\epsilon_{n+1}}{\epsilon_{n}^\phi}$$ The meaning of $\phi$ is well defined in the books, it says how fast the the result converge, but what is the meaning of the asymptotic error constant? For example when I used newton method for one root I got A around 1 (I tried one example), and when I used the extend newton method for some roots I got A below 1. So, what it's saying?

Answer 1

It is really not relevant in terms of the speed of convergence but, nevertheless, if you are comparing two sequences with the same rate of convergence $\phi$, a lower value of $A$ is better. In fact, $A=0$ would mean that the rate of convergence is higher than $\phi$.

Answer 2

Assume that from some point

$$\frac{\epsilon_{n+1}}{\epsilon_n^\phi}=A$$ exactly.

Then

$$\epsilon_{n+1}=A\epsilon_n^\phi$$

$$\epsilon_{n+2}=A(A\epsilon_n^\phi)^\phi=A^{\phi+1}\epsilon_n^{\phi^2}$$

$$\epsilon_{n+3}=A^{\phi^2+\phi+1}\epsilon_n^{\phi^3}$$

$$\epsilon_{n+4}=A^{\phi^3+\phi^2+\phi+1}\epsilon_n^{\phi^4}$$

$$\cdots$$

$$\epsilon_{n+k}=A^{(\phi^k-1)/(\phi-1)}\epsilon_n^{\phi^k}=A^{(\phi^k-1)/(\phi-1)}\epsilon_n^{\phi^k}.$$

This shows you how the factor $A$ influences the convergence.