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May
21
revised Total area for a natural nested set of convex polygons.
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May
21
revised Total area for a natural nested set of convex polygons.
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May
21
revised Total area for a natural nested set of convex polygons.
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May
21
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.
May
21
revised Total area for a natural nested set of convex polygons.
added 50 characters in body
May
21
answered Total area for a natural nested set of convex polygons.
May
20
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.
May
20
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.)
May
19
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.
May
19
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 $\mathbb R^2$.