Hello! I'm having trouble proving that there is and isomorphism between Dedeking Cuts and Cauchy Sequences, could someone take a view to this post? math.stackexchange.com/questions/4423954/… thank you :D
I know that ax+by=ab-a-b has no positive integer solutions. This is because (b-1,-1) is a solution of the equation so all solutions of the equation are given by x= b-1+m b, y= -1-ma, where m is any integer. Clearly, x and y can’t both be positive for any m.
Let $a$ and $b$ be two relatively prime positive integers. Prove that
$1)$ every sufficiently large positive integer $N$ can be written as a linear combination of $a$ and $b$ where $x$ and $y$ are both non-negative, i.e., there is an integer $N_0$ such that for all $N\ge N_0$, the equation $ax+by...
@copper.hat I didn't recognize the "p" in the 2nd one as momentum until after I posted it here - I thought it was going to turn out to be a mystery in need of a solution.
I need to combine these operations into one to get the given matrix $\boldsymbol{L}$. I need to use either a matrix product or a tensor product. Help, I can not find the operation.
@copper.hat I have three vectors and three matrices. I need to come up with a formula that would give a matrix whose diagonal elements would be the quadratic forms I have given. At first I thought that three vectors could be augmented into a matrix. Then I can use something like the formula: $\boldsymbol{L} \approx \boldsymbol{u}^T\boldsymbol{J}\boldsymbol{u}$
But the problem turned out to be that, firstly, my J matrices are different, and secondly, there are side off-diagonal elements in the result.
I thought that tensor operations would help me, but I haven't figured them out yet.
I need a sanity check: I'm comparing things in $H^2$ and $H^3$. Suppose you have a quadrilateral with all edges and angles being identical, and which you can tile $H^2$ with by packing 5 around every corner. Then you have a cuboid with all faces and dihedral angles being identical which you can tile $H^3$ with by packing 5 around every edge. Should the edge length of the quadrilateral in $H^2$ necessarily equal the edge length of the cuboid in $H^3$?
I feel like it should be, but my calculations are giving different results, which would imply that my calculations are wrong
Well, this is meant to be two entirely different cases. Quadrilaterals in $H^2$ and cuboids in $H^3$. However, I think I might have found an assumption I was making that clarifies why they would have different edge lengths, and I'm trying to articulate it.
Given a set of five cuboids around an edge and one of the two vertices that define that edge, consider the faces that have that vertex as one of their vertices (and which lack the other vertex of that edge as one of their vertices). Those face are not coplanar? Right?
@TedShifrin I am just trying to understand let us say we have rational function over some subvariety V $g = \frac{g_1}{g_2}$ then $div(g)_V = div(g_1)_V - div(g_2)_V = V(g_1) \cap V - div(g_2) \cap V$
I want to understand exactly the push forward of those cycles
My mom's in the hospital right now with Pneumonia. They aren't sure exactly what caused it, but it's certainly a complication from either her cancer or the chemo treatment
I think I've got the solution to my hyperbolic conundrum: Given a set of five cuboids around an edge, to relate the edge lengths to the edge lengths of five quadrilaterals around a vertex in $H^2$, we need to find a projection from $H^3$ to $H^2$ where the cuboids get flattened to quadrilaterals. That should be the projection onto the plane that cuts through the midpoints of the edges of the cuboids parallel to their shared side.
However, the edge lengths of these quadrilaterals after the projection are the geodesics of closest approach between the edges of the cuboids. So, the length of the edges of the cuboids themselves should be longer, right?
(Dangit, too slow to get that last typo. Should be "shared edge," not "shared side")
Honeybombs are weird in hyperbolic space. Like, in Euclidean space, you can have a cubic honeycomb, and if you take a vertex and surround it on all sides by cubes, you end up with a larger cube (edge length twice that of the smaller cubes). If you do the same with cuboids with dihedral angle $\frac{2\pi}{5}$ in hyperbolic space, you end up with twenty cubes around a vertex, and the space they fill up is a 60-sided polyhedron with none of the faces being coplanar
Now, this polyhedron with 60 quadrilateral faces. It, too, can form a honeycomb of hyperbolic space. I can't even begin to imagine what that would look like, though
