775 reputation
411
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location Chemnitz, Germany
age 27
visits member for 3 years, 3 months
seen 17 hours ago

Mar
2
comment How many squares actually ARE in this picture? Is this a trick question with no right answer?
Of course, what else. I wonder how this couldn't be a definite answer, a square is a square.
Feb
27
comment Dihedral angles between tetrahedron faces from triangles' angles at the tip
Since you were the first one to come up with a solution, it hurts not giving you the credit of acceptance. But I have to make a decision and achille's answer is more elaborate. But +1, of course.
Feb
26
comment Dihedral angles between tetrahedron faces from triangles' angles at the tip
+1 Man, of course, the normal vectors with some vector arithmetics, nothing 'bout spherical trigonometry, me stupid!
Feb
26
comment Dihedral angles between tetrahedron faces from triangles' angles at the tip
@SSumner Thanks for this link. "But I have no idea how it works" - Well, it has the formulas written on the site (though I'm also not able to directly derive this from spherical trigonometry, but at least I got a formula). Maybe somebody comes up with a nice answer showing the actual derivation of those formulas.
Feb
26
comment Dihedral angles between tetrahedron faces from triangles' angles at the tip
@SSummer Really? In the end the only thing I'm uncertain of is the bottom plane of the supposed tetrahedron and this plane's location shouldn't change the dihedral angles between the other three faces. The tip's angles should completely define the tip triangles' locations relative to each other, shouldn't it? If I'm wrong on this, you could make the counter-proof an answer.
Apr
24
comment What is the formula of the following?
I don't currently get what you mean by "the three orthogonal tangent planes". In each point of the ellipsoid you have one unique tangent plane. Or do you mean all triplets of pairwise orthogonal planes that are tangent to the ellipsoid? And of course I second the above three comments.
Apr
11
comment Intersection between 3D closed contour and 3D plane
@msotaquira A line-plane intersection is an extremely trivial operation. Come back when your contour has thousands of points to talk about computational expense.
Apr
10
comment Intersection between 3D closed contour and 3D plane
@msotaquira Well, then it's not a contour, is it? So you're rather asking for a good way to define a contour using those points, which is entirely dependent on the context and your needs. But using a simple piecewise linear curve, as Mark suggests in his answer (and AakashM assumed in his comment) would be a good start.
Mar
13
comment I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent
@t.b. I have never seen it anywhere before, but Ok, according to your link it seems apprpriate.
Mar
8
comment How to define sparseness of a vector?
Sorry if my comments are a bit confusing. It isn't the name sparseness that bothers me, it's the hard fact, that your above function (the one with the $\sqrt{k}$) is 1 for a sparse vector and 0 for a dense vector (no matter how you name it).
Mar
8
comment How to define sparseness of a vector?
I know, I just wanted to make clear, that your explanation as it stands is wrong, if Sparseness(X) is indeed defined as above.
Mar
8
comment How to define sparseness of a vector?
Isn't your sparseness function 1 for a sparse vector and 0 for a dense vector?
Feb
27
comment Why can we figure out the relationship between a point and a plane by a matrix's determinant?
@CChen No, the answer is not accepted, as you forgot to actually accept it.
Feb
16
comment $\beta_k$ for Conjugate Gradient Method
The idea of CG (in my limited understanding of theoretical numerical linear algebra) in contrast to gradient descent is to compute the current solution not only orthogonal to the current error, but also orthogonal to all the previous errors. And by using an appropriate basis (an A-orthogonal one) we get these nice properties of iterative computability and fast convergence. May the mathematicians forgive me if this generalized conceptual understanding is wrong.
Feb
16
comment $\beta_k$ for Conjugate Gradient Method
When taking the $r_k$ as basis, you don't get a minimal basis, meaning once you get to $r_{n+1}$ it cannot be a basis anymore, as this $r_{n+1}$ has to be linearly dependent on the previous $r$s (as the space is only $n$-dimensional at maximum). But when $A$-orthogonalizing the basis each time, you get a minimal basis of only $n$ vectors at maximum (making CG a direct method in theory).
Feb
16
comment $\beta_k$ for Conjugate Gradient Method
Of course it can. There are many bases for this sub-space, but only the $p$-basis is $A$-orthogonal, the $r$-basis isn't.
Feb
16
comment $\beta_k$ for Conjugate Gradient Method
$p_k$ and $r_k$ are not the same for $k>0$, because you don't just use the next orthogonal direction (which would be the gradient $r_k$). You take the gradient direction $r_k$ and orthogonalize this to all the previous $p_k$ (which span the current solution space), but with respect to $A$, making them conjugate with respect to $A$ and not just orthogonal (and this way also different from $r_k$).
Feb
10
comment Simplest equation for drawing a cube based on its center and/or other vertices
Well, a mass is not a length, but I guess your cube has a constant density of $1$, in which case the mass $m$ is indeed equal to the volume $V$ and you can determine the side length with a simple $s=\sqrt[3]{V}$. If not, then you need more information than just the mass to gain information about the cube's size.
Feb
9
comment Let $ \rho(P)$ be the spectral radius of $P$. Show $ \rho( \dfrac{P}{ \rho(P) + \epsilon } ) < 1 \text{ for all } \epsilon >0. $
Can it be you're missing a $c$ in there (first equation)? Otherwise I cannot believe $\lambda$ being an eigenvalue implies $c\lambda$ being an eigenvalue of the same matrix, too. With $\sigma$ you mean the spectrum, right?
Feb
7
comment matrix equation $(A-B)CA=B$
It may be simple, but why does $Q(CA+I)=A$ make $Q$ invertible?