New answers tagged

0

Not an answer, but a related result that I wanted to share. Note that $k$-distance-transitivity is generalized by $k$-walk-regularity. In Camara, van Dam, Koolen, Park: "Geometric aspects of 2-walk-regular graphs" the authors prove the following (see Corollary 6.4): Theorem. Let $G$ be a $k$-walk-regular graph of degree $d$ with an eigenvalue $\...


1

This is not even true for real-valued functions. The standard counterexample is the Cantor function, which is differentiable a.e. with derivative $0$, but is not constant as any absolutely continuous function with this property would be.


2

I read your paper (Sums of squared distances between points on a Unit sphere 2020 arXiv), but I must inform you that your theorem 2.1 was already established by Tom M. Apostol in a paper in Math Monthly 2003 (Sums of squares of distances in m-space) in a more general way (weighted).


0

Assuming you can easily compute the infimum and supremum of the density function $f$ on a given square with axis-aligned sides, you can use a quadtree to perform an approximate sampling. Let's assume that $f$ is bounded. Work in 2D Cartesian coordinates and write $S_{l, b, \delta}$ for the axis-aligned square containing all points $(x,y)$ with $l \leq x < ...


1

We are ultimately looking at the complexity of the multivalued function $\mathrm{MidPoint}_\mathbf{X} : \mathbf{X} \times \mathbf{X} \rightrightarrows \mathbf{X}$ assigning some midpoint to the points here. This includes what choice functions there are, but need not be limited to it. The framework to study the complexity of such operations is Weihrauch ...


2

A circular disc cannot be captured. Likely the same idea can be used to show that any convex body of revolution cannot be captured. It can be proved using a small variation of the idea as in the original answer. It is sufficient to show that infinitesimal Möbius tranform $m$ of the disc can shorten the wrapping length while preserving the crossing pattern. ...


9

We will use the Kuratowski–Ryll-Nardzweski selection theorem: Let $(\Omega, \mathscr{F})$ be a measurable space. Let $E$ be a Polish space. Let $\Gamma$ be a set-valued function from $\Omega$ to $E$; that is, for each $\omega \in \Omega$, let a set $\Gamma(\omega) \subseteq E$ be given. Assume that, for all $\omega \in \Omega$, the set $\Gamma(\omega)$ is ...


7

If $(X,d)$ is a complete metric space with the algebraic midpoint property (i.e. for all $x$ and $y$ in $X$, there exists $z\in X$ such that $d(x,z)=d(y,z)=d(x,y)/2$) then $X$ is a path metric space. Indeed, for all $x,y\in X$ one can iteratively construct a map $\gamma$ from $[0,1]\cap\mathbb D$ to $X$ such that $d(\gamma(s),\gamma(t))=|t-s|$, and extend it ...


5

Maybe the unit circle embedded in the euclidean plane is an example of a space that has several measurable midpoints structures but no continuous such structure? Let us choose as the middle point of two points which are not on a diameter the point in the middle of the shortest arc connecting the two points. When two points are on a diameter, we may choose ...


2

Let us construct an example of a measurable midpoint space that is not a continuous midpoint space. The idea is to create a "jump" of midpoints somewhere. One way to do that is to consider a slit rectangle, e.g. $$ \tilde E = [0,1]\times[0,1]\setminus \{\frac12\}\times(0,1) $$ endowed with the length metric induced by the canonical euclidean scalar ...


3

The following general result is described in the answer I posted on math.stackexchange here: if $V_1$ and $V_2$ are finite-dimensional vector spaces of equal dimension over an arbitrary field and they are both equipped with nondegenerate bilinear forms $B_1$ and $B_2$, then a function $\sigma \colon V_1 \rightarrow V_2$ such that $B_1(v,w) = B_2(\sigma(v),\...


1

Loosely speaking, if you are standing at the origin then there are only three directions you can travel. So the space of directions $S_o$ is only 3 points. Taking the product of the space of directions with $[0,\infty)$ and identifying $S_o \times \{0\}$ to a point gives you the tangent cone, which in this case is isometric to your starting space. If you are ...


7

The answer is No, there are no other such polytopes, as I was able to show in this recent preprint. Theorem. In dimension $d\ge 4$, an edge-transitive polytope is vertex-transitive. The idea is as follows: first, show that every edge-transitive polytope $P$ that is not vertex-transitive has the following three properties: all edges of $P$ are of the same ...


0

There are two phrases that may help in your search: Point-set registration. The link is to a (long) Wikipedia article, which includes "rigid registration," which seems closest to your case. Geometric shape matching. For example: Alt, Helmut, and Leonidas J. Guibas. "Discrete geometric shapes: Matching, interpolation, and approximation." ...


2

I suspect not, at least for regular polygons. (For nonregular polygons, there are orientation issues which I believe won't be solved by rubber bands.) Let me illustrate with squares. Consider 8 squares in a three by three arrangement with a central hole. This configuration by itself does not tile when you apply a rubber band, but if you have a larger ...


4

Without loss of generality assume that $O=(0,0)$ and the ABCD circle has radius 1. Then $$\begin{cases} A=(\cos \alpha,\sin\alpha),\\ B = (\cos\beta,\sin\beta),\\ C = (\cos\gamma,\sin\gamma),\\ D = (\cos\delta,\sin\delta),\\ AB = 2\sin\frac{\alpha-\beta}2,\\ BC = 2\sin\frac{\beta-\gamma}2,\\ CD = 2\sin\frac{\gamma-\delta}2,\\ AD = 2\sin\frac{\alpha-\delta}...


1

One more reference (making no smoothness assumptions on $\partial \Omega$) is G. Fast. Area of a generalized circle as a function of its radius. I, II (in Russian). Fund. Math. 46:137--163, 1959. I tried to find more modern results than this one and those cited by Ivan Izmestiev. I expected to get an answer from D. Hug, G. Last and W. Weil (2004) A local ...


1

Your formula for area (or I should say volume form), is always true up to within an error of size $o(r)$. The geometry (boundary, curvature, etc.), only comes into play when you try to write down the second-order correction term. In fact you can define the boundary measure (aka perimeter) of sets that don't have boundary using that formula. It is the so-...


4

The reflection of a point $p$ in a point $u$ is given (if we identify points with vectors) by $2u-p$. So the composition of two such reflections, say about points $u$ and $v$, is given by the translation $p\mapsto2v-(2u-p)=p+2(v-u)$. By induction, the composition of reflections in $u_1,v_1,u_2,v_2,\dots u_n,v_n$ is the translation $p\mapsto p+2(V-U)$ where $...


5

We describe the $B_i$'s vectorially. For simplicity I write $AB$ instead of $\vec{AB}$ $$ OB_1=OP+2PA_1= OP+2(OA_1-OP)=2OA_1-OP$$ $$OB_2= OB_1+2(OA_2-OB_1)= 2OA_2-OB_1=2OA_2 -2OA_1+OP $$ $$OB_3=OB_2+ 2(OA_3-OB_2)= 2OA_3-OB_2=2OA_3-2OA_2+2OA_1-OP.$$ The midpoint $M$ of $PB_3$ is given by $$OM=\frac{1}{2}(OP+OB_3)=OA_3-OA_2+OA_1. $$ Same computation ...


1

Call a quaternion whose scalar part is zero a vector quaternion. We shall denote the vector quaternions as $\mathbb R^3$. Given $q = w + xi + yj + zk$, we shall define $q^*$ (called the "conjugate" of $q$) to be $w - xi - yj - zk$. If $q$ is a unit quaternion, then $v \in \mathbb R^3\mapsto qvq^*$ is a rotation. All rotations about the origin in 3D can be ...


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As it is a version of set-cover, the problem is NP-hard. The following is a special case: (1) Chepoi, Victor, and Bertrand Estellon. "Packing and covering $\delta$-hyperbolic spaces by balls." In Approximation, Randomization, and Combinatorial Optimization. Algorithms and Techniques, pp. 59-73. Springer, Berlin, Heidelberg, 2007. Springer link. "In ...


10

I have looked at Goldberg's paper referenced by J. J. Castro in his excellent answer. It turns out that there is a simpler (and more general way) to generate Goldberg's non-circular solutions, so I thought that I would just mention that. The idea is to consider the 'rotor' (Goldberg's term) as fixed and as the envelope of a circle in periodic motion. It's ...


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