19

TL;DR: The continuous operator exhibits this behavior, any faithful discretization will just inherit it. Deeper cut: If you look at the spectrum (eigenpairs) of the continuous operator $\frac{d^2}{dx^2}$, the eigenvectors are trigonometric functions of the form $\cos(kx)$ and $\sin(kx)$, with eigenvalue $k^2$. Low frequency functions (ie constant or very ...


10

Jean-Michel Muller, et. al., "Handbook of Floating-Point Arithmetic 2nd ed.", Birkhäuser 2018, gives the following example due to Muller, specifically constructed to deliver incorrect results with floating-point evaluation: $$ {u_{0} = 2,\\ u_{1} = -4,\\ u_{n} = 111 - \frac{1130}{u_{n-1}} + \frac{3000}{u_{n-1}u_{n-2}},\>\>\>\>n \ge 2.} $$ ...


6

The previous and next IEEE machine numbers to $\alpha_j$ are at a distance $\approx |\alpha_j| \varepsilon_{mach}$ from each other; hence $fl(\alpha_j)$ (the closest machine number to $\alpha$) is at a distance at most $|\alpha_j| \varepsilon_{mach}$ from $\alpha_j$, but that's just an upper bound: if $\alpha=0$ or $\alpha=1$, for instance, then it is ...


6

In my opinion, an example could be the calculation of the minimal solution of a three term recurrence relation (TTRR) $$y_{n+1} +a_{n}y_{n}+b_{n}y_{n-1} = 0, \quad n=1,2,3,\ldots$$. For example, the Bessel $J_{n}(x)$ function for a fixed $x$ satisfies the TTRR $$y_{n+1}-\frac{2n}{x}y_{n}+y_{n-1} = 0$$ Suppose you know $y_{0} = J_{0}(1)$ and $y_{1} = J_{1}(...


2

1. For a matrix $A$ with distinct eigenvalues, adding a perturbation $\delta A$ results [1] in a change to eigenvalues of magnitude (to first order) $$ \delta\lambda_i = (X^{-1}\delta A X)_{ii}, $$ so $|\delta \lambda_i| \leq \kappa(X)\|\delta A\|$. The change in condition number $\kappa(A)$ is then given by $$ \delta(\kappa(A)) = \kappa(A)\big(\delta(\log \...


2

For sparse matrices $A$, that are a discretized version of operators in PDEs in FE, FV, or FD, you do know your sparsity pattern before you compute the actual entries. So, you usually compute matrix elements of $A$ that are not zeros according to the sparsity pattern do not do anything with your known zeros. In your question, it seems that you want to go ...


2

It might help to start with a concrete example of a function, like $f(x,y)=10^9 x^2+y^2$ (the example they use on p.26). I'm trying to guess what went wrong with your analysis, because you haven't given much information. I am mostly struggling with cross-terms of hessian. Consider $$f_1(x,y) = 10^9 x^2+y^2$$ and its rotation by $\theta=\frac\pi4$ $$f_2(x,...


1

I don't think your own answer is correct, though my method somewhat differs in so far as I use a different norm than you - the Euclidean norm. My result: The relative condition for the sum of two numbers $x+y$ is $$\kappa_R = \dfrac{\sqrt{2x^2+2y^2}}{|x+y|}.$$ The absolute condition is $$\kappa_A = \sqrt 2\approx 1.41.$$ Hence the sum is very ...


1

The mistake I made is that the Jacobian in the infinity norm is equal to 1, not $n$. As such, the relative condition number of the sum is: $$\kappa=1$$ and as such is always well-conditioned, no matter what magnitudes the values are.


1

If I understand your question correctly you are looking for a factorisation $$\tilde{C}^T \tilde{C} = C^TMC,$$ where $M$ is symmetric positive definite. Moreover you have at hand $M$ and $B=MC$. For $M$ you can compute Cholesky, LDLT, and eigenvalue decompositions: $$M = L_1 L_1^T = L_2 D_2 L_2^T = Q \Lambda Q^T,$$ where $L$ are lower triangular, $Q$ ...


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