if $\lim_{k\to \infty} x^k = x^*$ then the error by our approximation $x^k$ is
$$e_{k} = ||x^{k}-x^*||$$
Quadratic convergence means $$e_{k+1}\le ae_k^2$$
where $a>0$
according to my book. I don't see why this converges. Maybe it should be:
$$\lim_{k\to\infty} \frac{e_{k+1}}{e_k^2}\le a$$
for some $a>0$
I must prove that it implies superlinear convergence:
$$\lim_{k\to \infty} \frac{e_{k+1}}{e_k}=0$$
I don't see why quadratic implies superlinear. I can see, however, that superlinear implies quadratic, I think. Because $\lim \frac{e_{k+1}}{e_k^2} = \lim \frac{e_{k+1}}{e_k}\frac{1}{e_k}$ which is the limit of something going to $0$ and something bounded.
Where is the error in my argument and how to prove quadratic convergence implies superlinear convergence?
Best Answer
Note that $e_k \to 0$. If $(x^k)$ is quadratic convergent, then $e_{k+1} \leqslant ae_k^2$, or $$ 0 \leqslant \frac {e_{k+1}}{e_k} \leqslant ae_k \to 0 [n\to \infty]. $$ Now the conclusion follows the squeezing theorem.