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seen Nov 6 '13 at 3:42

May
21
comment Lie and Weierstrass' visualization of complex functions
Over the weekend, I thought up a construction similar to yours. To a complex number $x+iy$ in the functions argument I associate the line connecting $(0,0,-1)$ and $(x,y,0)$. An upward unit vector in the direction of this line acts like your $v$. Like you, I map this line to a member of its orthogonal complement of parallel lines. I orient this complementary space using a vector in the direction of the real axis original complex plane (instead of $n$). But I also have difficulty seeing how this collection of lines in three space provides insight into the behavior of a complex function.
May
12
comment Lie and Weierstrass' visualization of complex functions
I am with you as far as you went, but what then? Something involving a map from the Riemann sphere, as lines, to all the lines in 3-space? Nothing I thought of made much sense. Certainly nothing provided any insight into the function being modeled.
May
12
comment Lie and Weierstrass' visualization of complex functions
@deoxygerbe The quote is on page 41 in either the third (1920) or fourth (1927) edition — the first section of the third chapter.
Apr
22
comment Enigma : of Wizards, Dwarves and Hats
I believe the problem statement precludes any communication once the dwarves decide on a strategy or, at least, that is the intention behind the questioners P.S. section.
Apr
22
comment Enigma : of Wizards, Dwarves and Hats
You may want to add that the dwarves need to have absolutely perfect eye sight with infinite resolution and an uncountably infinite capacity to process what they see.
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
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
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
13
comment Linear Algebra of Symmetric Sums: a converse question
I found a straightforward way to calculate the constant term from the from the powers sums, but I have not found a good way to calculate the discriminant from the power sums. Is the only way to calculate the all the coefficients of the polynomial and then find the determinant of the Sylvester matrix?
Mar
12
comment Do equal mean and equal moment imply equal distribution?
@user7815: This paper http://ramanujan.math.trinity.edu/wtrench/research/papers/RP-112.pdf by William F. Trench appears relevant.
Mar
12
comment Linear Algebra of Symmetric Sums: a converse question
Thank you for your answer. I think that you have resolved the "some of $z_j$'s are zero" issue. There is an algebra error in your $n=2$ non-distinct discussion. The criteria should be $$\frac{1}{2}\mathcal{Z_1}^2=\mathcal{Z_2}.$$
Mar
12
comment Center of gravity of a self intersecting irregular polygon
@mixkat: Given your description of the problem, finding the centroid of the polygon your group forms seems like overkill. Treat each person like a point mass and find the center of mass of your collection of points as: $$\left(\frac{\sum_{i=0}^n x_i}{n}, \frac{\sum_{i=0}^n y_i}{n}\right).$$
Mar
12
comment Linear Algebra of Symmetric Sums: a converse question
My question is: given only the $\mathcal{Z_k}$'s and that $x=(1,1, \dots, 1)$ can I tell if the matrix $M$ is singular or not?
Mar
10
comment Equidecomposability of a Cube into 6 Trirectangular Tetrahedra
Just making sure everyone is aware: the easiest generalization of the 2-dimesional splitting of a square into two right triangles comes out of noting that you can split the square region ($0<x,y<1$) into two congruent triangular regions: $x>y$ and $y>x$. This generalizes to 3 dimensions by splitting the cubic region ($0<x,y,z<1$) into 6 congruent tetrahedral regions $x<y<z$, $y<x<z$, $x<z<y$, $z<x<y$, $y<z<x$, and $z<y<x$. This easily generalizes into higher dimensions.
Mar
9
comment From baby Hartshorne: showing the exterior of a circle is segment connected
Again the only access I have to Hartshorne is the Google excerpts, but I know that Hilbert created an arithmetic of congruent segments (Hilbert's Foundations of Geometry see pages 15-16 and 23-36). I assume he applied that arithmetic to angle measures using Archimedes' approach for successively better estimations of arc-length.