Reputation
22,919
Next tag badge:
84/100 score
19/20 answers
Badges
3 36 85
Newest
 Guru
Impact
~425k people reached

14h
comment Does a fluid with $0$ divergence have $0$ density?
If $\mathrm d\rho/\mathrm dt=0$, it doesn't mean that $\rho=0$. It just means that $\rho$ is constant. (Everyone forgets the constant of integration, grumble grumble...)
1d
comment Way to measure the similarity/difference of 2D point clouds
If the curves are noisy, that could still throw the arc length parametrizations out of sync (imagine one curve is noisier near the beginning and the other is noisier near the end, and you get the same situation as my first comment). Which is why I suggested the Hausdorff distance, which is independent of parametrization.
1d
comment Way to measure the similarity/difference of 2D point clouds
This is halfway to a good idea. But if the first point set has a lot of points near the beginning of the curve and the second has lots of points near the end of the same curve, your metric will still give a large dissimilarity. What you should do, once you have a notion of the curves represented by each point set, is to measure the geometrical dissimilarity between the curves themselves, for example using the Hausdorff distance.
1d
comment Total area for a natural nested set of convex polygons.
Whoops, that's not quite true. When $n$ is large and odd, $P^*$ forms two simple polygons of area close to $A$, so $A^*\approx A$, or a single "doubly-wrapped" polygon whose central area is double-counted, and again $A^*\approx A$. I'll edit my answer.
1d
comment Why $ (A\vec{x})'A \vec{x} = \vec{0}$ implies that $A\vec{x} = \vec{0}$
You still have $A^TAx=x^TA^TAx$, which is also incorrect.
1d
comment Why $ (A\vec{x})'A \vec{x} = \vec{0}$ implies that $A\vec{x} = \vec{0}$
The argument as written is incorrect. It is true that if $x^TA^TAx=0$ then $Ax=0$, but it is not true that $x^TA^TAx=Ax$. (Nor that $A^TAx=x^TA^TAx$, as written in the immediately previous line.)
2d
comment Identification of a quadrilateral as a trapezoid, rectangle, or square
@Daniel: Even if that were the case, do we expect students taking the test to answer the question as given, or do we expect them to somehow divine the intent of the question setters? ...On second thought, given the quality of standardized testing in many places, don't answer that.
2d
comment mathematical symbol for vector appending
This is a natural application of block matrix notation. If we can write $$\begin{bmatrix}\mathbf A&\mathbf b\\\mathbf b^T&c\end{bmatrix}$$ where $\mathbf A$, $\mathbf b$, and $c$ are a matrix, a column vector, and a scalar respectively, then surely we can write $$\begin{bmatrix}\mathbf v\\w\end{bmatrix}$$ to denote $$\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\\w\end{bmatrix}$$ (and we do).
May
19
comment The difference between a basis of vectors vs functions
Your space of functions is not a subspace of $\mathbb R^2$, because none of the functions in it are elements of $\mathbb R^2$. The $xy$-plane is not a subspace of $\mathbb R^2$ either -- maybe that was a poorly chosen example, consider the subspace $x\begin{bmatrix}1\\2\\0\end{bmatrix} + y\begin{bmatrix}3\\-1\\-2\end{bmatrix}$ instead. The point is that you should stop thinking of two-dimensional vector spaces as being inherently tied to $\mathbb R^2$.
May
19
comment The difference between a basis of vectors vs functions
The space of functions spanned by $\sin(t)$ and $\cos(t)$ is simply the set of all functions of the form $x\sin(t)+y\sin(t)$. You don't need the extra $t$ component. Perhaps you're thinking that any two-dimensional vector space must be written with two components in an array, but that's not true. For example, the $xy$-plane is a two-dimensional subspace of 3D space, and any point on it is of the form $x\begin{bmatrix}1\\0\\0\end{bmatrix} + y\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}x\\y\\0\end{bmatrix}$.
May
18
comment Fourier transform of a product of two rect functions
The Fourier transform of $f(x)g(x)$ is $\hat f(x)*\hat g(x)$, but the Fourier transform of $f(x)g(y)$ is just $\hat f(x)\hat g(y)$.
May
18
comment Is this a spontaneous symmetry-breaking?
It sure looks spontaneous to me, because there is no explicit term breaking the symmetry, but then all I know is the Wikipedia article I just looked up. You might find more people who know about symmetry breaking at the physics site.
May
18
comment What is the average width of a given tetrahedron?
A fantastic answer to a not-so-promising question!
May
17
comment “Vectors aren't really numbers” - how sound is that statement?
What exactly is a number?
May
17
comment Proving that the matrix is positive definite
@Davide: Why did you put the "P" in the title into math mode?
May
17
comment Why is Cumulative “Density” wrong?
The term density in general refers to the amount of something per unit length/area/volume, e.g. mass density, charge density, etc. The probability density function tells you, literally, the density of probability. It is something you integrate over a region to obtain the actual probability. The cumulative distribution function, on the other hand, tells you the actual integrated probability between $-\infty$ and the point in question, so it is not a density: it is not something you can integrate over a region to obtain a sensible quantity.
May
17
comment How is the vector space of abstract “tuples” isomorphic to vector space of $n \times 1$ or $1 \times n $ matrices?
The point is not whether the proof is hard or easy, the point is that your answer makes it sound like the proof is not necessary, which is misleading. Even in the comments you implied that any bijection between vector spaces is an isomorphism, which is not true.
May
17
comment How is the vector space of abstract “tuples” isomorphic to vector space of $n \times 1$ or $1 \times n $ matrices?
You also have to show that the bijection is a homomorphism. It is not enough that a bijection exists and a homomorphism exists -- consider $\mathbb R$ and $\mathbb R^2$.
May
16
comment Is there a Continuous Multinomial Distribution??
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
May
15
comment Hexagon packing in a circle
Vaguely related: Gauss circle problem