# Tag Info

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There are two principles at play here: a mathematical principle that favors hexagonal networks, and a physical principle that favors a network with straight walls. The mathematical principle that prefers hexagonal planar networks is Euler's theorem applied to the two-torus $\mathbb{T}^2$ (to avoid boundary effects), $$V-E+F=0,$$ with $V$ the number of ...

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There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]: Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent: $p$ is a sum of two squares in $\mathbf Z[x]$; $p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$; Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two ...

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11

There is a (nearly) trivial linear-time lower bound, because it takes linear time to read in the input. (I say this is nearly trivial because you do have to argue that it is necessary to examine at least a constant fraction of the input to get the right answer.) Some slightly super-linear time bounds are known, under certain assumptions about the ...

11

To answer on methods applicable here (and elaborate on comments I made). The most promising is to use a surprisingly little-known theorem that says that the discriminant $D$ of a symmetric $n\times n$ matrix $A=(a_{ij})$ with eigenvalues $\lambda_1,\dots,\lambda_n$, i.e. $$D_A=\prod_{1\leq i<j\leq n} (\lambda_i-\lambda_j)^2,$$ is a sum of squares in the ...

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The rigorous mathematical context is stability. A straight rope in either tension or compression is a valid solution of the underlying PDE, but in compression this solution is unstable, so it cannot be realized in practice.

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The Carleton College library has a copy of the Kushner book. Here's the theorem: Theorem 8 Let* $P\gt0, C\ge0$ and $$EA_n'PA_n-P=-C.\ \ \ (8.24)$$ Then $EX_n'CX_n\rightarrow0$ and $X_n'CX_n\rightarrow0$ w.p.l. Also $$P_x(\sup_{\infty\gt n\ge0} X_n'PX_n \ge \lambda) \le {x'Px\over\lambda}.$$ Hence, the ...

10

First of all, there is a general theory (due to Chebyshev) on the best uniform approximation of ANY continuoius function $f$ by polynomials of degree at most $d$ on an interval. It describes the polynomial of the best approximation, which is unique. In Chebyshev's polynomials, $f=x^n$ and $d=n-1$. You are asking about $f=x^n$ and some given $d<n$. I do ...

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Isn't it just the 2d sphere packing? If one assumes that the larvae needs a disc of fixed radius to grow up to an adult form and that the bees want to have as many cells as possible then the hexagonal lattice is the optimal one.

10

Write $x = 1-X$, $y=1-Y$, $z=1-Z$. Then the inequality reduces to $$2XYZ \leq X^2 + Y^2 + Z^2$$ for $X, Y, Z \leq 1$. If $X, Y, Z < 0$ then the inequality is trivial, since LHS < 0. Otherwise suppose $X \in [0, 1]$ wlog. Then $$\text{RHS} - \text{LHS} = X^2 + (1-X^2)Y^2 + (Z-XY)^2 \geq 0.$$ Equality holds iff $X = Y = Z = 0$.

9

First, let's assume $a=1$; for other values we can scale a solution with $\sqrt{a}$. So we want to minimize $H=\sum_{i,j} (1-\|x_i-x_j\|^2)^2$. I globally optimized the problem numerically for $n=4$ and $n=5$ and obtained as solutions for $n=4$, the square with side length $\sqrt{\frac{2}{3}}\approx0.8165$, giving $H_4=\frac{2}{3}$ for $n=5$, the ...

9

Since this is my main area of research, let me attempt an answer (which inevitably got quite long!). The short answer is that I do not know of any specific instances where topological complexity has been used to solve robotics problems, but I know that roboticists are interested in the concept, and am hopeful that such instances may occur in the future. Let ...

9

That is highly unlikely. Take any such system $S$ of permutations. Take any configuration of points obtained by sampling and note that we can also get it with the points re-enumerated in any way. Now let us look at how many reasonably short path enumerations are there at all. For each short path, there is a sequence of integers $a_1,\dots,a_{n-1}$ with sum $\... 9 By density, it is enough to prove the property when$A$is positive definite. Then $$A(I+BA)^{-1}=A^{1/2}(I+A^{1/2}BA^{1/2})^{-1}A^{1/2}$$ is congruent to$(I+A^{1/2}BA^{1/2})^{-1}$, which itself is positive definite because$I+A^{1/2}BA^{1/2}\succeq I$. 9 Zebras win for all$N$. I didn't realize Lawrence's answer in the source is actually sound (or so I think, when I really took some time to read it through this morning). Below I basically adopt Lawrence's strategy for$N$, with schematic drawings to make the argument easier to follow. The following is a winning starting position for the zebras. where$a$... 9 Here is a classic article by L. Fejes Toth on this subject. https://projecteuclid.org/euclid.bams/1183526078 9 Here is a paragraph of THE LIFE OF THE BEE (1901) By Maurice Maeterlinck: "There are only," says Dr. Reid, "three possible figures of the cells which can make them all equal and similar, without any useless interstices. These are the equilateral triangle, the square, and the regular hexagon. Mathematicians know that there is not a fourth way ... 9 It can be$O(n^{\frac32})$for$a\ge 1$if the sets$A_i$correspond to the$p^2$points of a smooth surface in an appropriate surface in a 3-dimensional space over$\mathbb F_p$and your points are the$p^3$general position planes, with$p^2$planes through each point. There are no 3 collinear points if the surface is chosen appropriately, so for any 3 ... 8 Since you know already that the optimal$a_i$have$2a_i + 1 = x_i = \pm 1$and the roots of$P'_{n-1}$, the calculation of$V_n$comes down to the discriminant of$P'_{n-1}$, its leading coefficient, and its values at$\pm 1$, all of which are available in closed form via formulas for Jacobia polynomials (since$P'_{n-1}$is a multiple of$P^{(1,1)}_{n-2}\$...

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