I have a function $f(x)$ and I calculated iterations until I got the root.
I have proven that it has quadratic convergence.
However, I am now asked to estimate the asymptotic convergence factor.
This is the factor K that relates the error between each iteration, is it not?
$K= \max \frac{f''(x)}{2 \min f'(x)}$?
When doing so, however, I got a K=60, which would mean that with each iteration, the error would greatly increasy which doesn't make sense given that it converges so quickly.
On
Suppose $x_{n}\rightarrow x^{\star}$. If we can write $$ \lim_{n\rightarrow\infty}\frac{\left|x_{n+1}-x^{\star}\right|}{\left|x_{n}-x^{\star}\right|^{r}}=L $$ for $0<L<\infty$ and $r\geq1$, we call $r$ the rate of convergence. An approximation of this is given by $$ r\approx\frac{\log\left|\left(x_{n+1}-x_{n}\right)/\left(x_{n}-x_{n-1}\right)\right|}{\log\left|\left(x_{n}-x_{n-1}\right)/\left(x_{n-1}-x_{n-2}\right)\right|}. $$ See this document for a gentle introduction.
If you have shown $f(x)$ converges quadratically, then you have shown that there exists a finite $K$ such that $$ \lim_{n\to\infty} \frac{|\varepsilon_{n+1}|}{\varepsilon_n^2} = $K$ $$ Here, $\varepsilon_n$ stands for the error in the $n$-th guess.
If you have shown that it converges quadratically but no faster, then your $K > 0$.
$K$ is the asymptotic convergence factor.
Example: Say $f(x) = x^2-1$.
Then if $x_k = 1+\varepsilon_k$, $$ x_{k+1} = 1+\varepsilon_k-\frac{(1+\varepsilon_k)^2-1}{2(1+\varepsilon_k)} =1+\frac12\varepsilon_k^2+O(\varepsilon_k^3)\\ \varepsilon_{k+1}=\frac12\varepsilon_k^2+O(\varepsilon_k^3) \\ \lim_{n\to\infty} \frac{|\varepsilon_{n+1}|}{\varepsilon_n^2} = \frac12 $$ so for this function, NR converges quadratically with an asymptotic convergence factor of $\frac12$.
Of course, for this simple function you can get the closed form value of the $n$-th guess given some starting guess, as an expression involving hyperbolic tangents and arctangetns and $2^n$.