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Mar
24
comment Chebyshev center = center of mass?
@lhf: It seems as though there will be lots of diverse and sundry polygons that work. Take a circle and three distinct points on that circle. Imagine three sides of a polygon lying on the tangent lines for the three points. Now connect those three side by inserting any number of sides between them. You need not worry about symmetry. Just make sure that the weight of the sections of the polygon between the three points of tangency balance so that the centroid is the center of the circle. I'd be more hopeful of an interesting result if the circumcenter and centroid were coincident.
Mar
24
comment Dirichlet forms
@Glassjawed As Didier noted, the gradient is defined on edges. The $\frac{1}{2} $ prevents one from double counting the edges when one sums over the vertices.
Mar
23
comment Chebyshev center = center of mass?
Would not a prism formed from a rotationally symmetric polygon work as well provided it was just long enough to make the incenter unique?
Mar
23
comment Dirichlet forms
You may want to look at Qiaochu Yuan's answer to this question on the discrete Laplacian. He describes the discrete Lapacian on a regular lattice. You appear to be looking at what amounts to a weighted graph. In that case, you need to know whether $x$ and $y$ are edge connected and how 'strongly'. You need $K(x,y)$ for that.
Mar
22
comment $2$ equations with $4$ variables
It appears that you also need to make sure $b>0$ is accounted for in your solution area. Should it be $\frac{(a-1)^2}{4} \gt b \gt \max(0, a-2)$ and $3 \gt a \gt 1$?
Mar
22
awarded  Disciplined
Mar
22
awarded  Organizer
Mar
19
revised Isosceles trapezoid
Struck an erroneous statement
Mar
19
revised Isosceles trapezoid
edited body
Mar
19
revised Isosceles trapezoid
added 122 characters in body; deleted 3 characters in body
Mar
19
revised Isosceles trapezoid
added 201 characters in body
Mar
19
answered Isosceles trapezoid
Mar
18
comment compute the fraction of products
I do not follow this implication:$$G(z) \frac{1+z}{\lambda z} = G(z/2) \Longrightarrow G(z) = \lambda \frac{z}{(1+z)(1-\frac{z}{2})}.$$ Can you clarify what happened?
Mar
18
comment The discrete Fourier transform of a Dirichlet charachter
NIST Digital Library of Mathematical Functions has a section on Periodic Number-Theoretic Functions that might be helpful.
Mar
17
awarded  Altruist
Mar
16
awarded  Investor
Mar
15
comment Estimating the Primarithm
Thank you. So roughly, since $n < pog(n^n)$ $$\frac{\log{n}}{\log \log{n}} < pog(n) < 1.5 \frac{\log{n}}{\log \log{n}}$$.
Mar
15
accepted Estimating the Primarithm
Mar
15
asked Estimating the Primarithm
Mar
13
accepted Linear Algebra of Symmetric Sums: a converse question