Numerical analysis question -- what am I being asked here?

I'm new to numerical analysis, and I'm working on the following. I'm given a discretization form of a differential equation:

$$ -\cfrac{p}{h^2} (U_{i+1} −2U_i + U_{i−1}) + qU_i = λ^hU_i, \quad 1 \leq i \leq N $$

with conditions: $U_0 = 0, U_{N-1}=U_{N+1}$. After writing this in matrix form, I get an eigenvalue problem $Au = \lambda u$. Then I'm asked to do the following:

Numerically demonstrate that the smallest eigenvalue converges to the true smallest eigenvalue in the second order by taking $p = 1, q = 5$, and $N = 2^n$, $n = 4,5,6,7,8,9$. You may plot the convergence order as a curve of slope 2

I know that I can find the smallest eigenvalue using the inverse power method, but I'm confused about what it means to plot "the convergence order as a curve of slope 2."

If I let $x^*$ denote the true smallest eigenvalue and $x_i$ the computed eigenvalue at the $i$-th step of the inverse power method, do I need to show that the limit

$$ \lim_{n\rightarrow \infty} \cfrac{|x_{n+1}-x^*|}{|x_{n}-x^*|^2}$$

converges?

Thanks!

Solution 1:

No, you are not required to complete a mathematical proof. Since you know the target value, the error $e_n = x^*- x_{n}$ can be recorded. If you have quadratic convergence, then as you say, $e_{n+1} / e_{n}^2$ should tend to a constant, say, $c$. It follows that $$e_{n+1} \approx c e_{n}^2$$ for $n$ sufficiently large. It follows that $$\log(|e_{n+1}|) \approx \log |c| + 2 \log(|e_n|),$$ so if you plot $\log(|e_{n+1}|)$ as a function of $\log(|e_n|)$ you should see a straight line with slope $2$.

However, there are significant practical limitations. It is exceedingly likely that you run out of digits before $n$ is sufficiently large. Typically, you are lucky to get three points on a nearly straight line and this is about as good as it gets.