Timeline for Is it known? A sum over lattice parallelograms of area one is equal to $\pi$
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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Sep 21 at 13:54 | comment | added | Nikita Kalinin | @DavidESpeyer yes, exactly! | |
Sep 21 at 12:27 | comment | added | David E Speyer | In fact, $F(x,y)$ is the area of the triangle with vertices $0$, $\phi(x)$, $\phi(y)$. So $F(x,x+y) + F(x+y,y) - F(x,y)$ is the area of our triangle with vertices $\phi(x)$, $\phi(y)$, $\phi(x+y)$. It makes sense that the area of the triangle with vertices $0$, $\phi(x)$, $\phi(y)$ is $\tfrac{\sin(2 \theta)}{2}$ because $\phi$ doubles angles: If $z = r (\cos \alpha, \sin \alpha)$ then $\phi(z) = (\cos (2 \alpha), \sin (2 \alpha))$. | |
Sep 21 at 12:20 | comment | added | David E Speyer | Nice! If you write $F(x,y) = \frac{(x \cdot y) (x \times y)}{|x|^2 |y|^2}$, then you can see that $F(x,y) = \sin(\theta) \cos(\theta) = \sin(2 \theta)/2$, where $\theta$ is the angle between $\vec{x}$ and $\vec{y}$. | |
Sep 21 at 6:28 | comment | added | Nikita Kalinin | Let us consider $F(x,y) = \frac{x\cdot y}{|x^2|\cdot |y|^2}$. Then, $F(x,y)-F(x+y,y)-F(x,x+y)$ is equal to our term. Telescope it, using that for big x,y we have $F(x,y)\approx $ the angle between x and y | |
Sep 19 at 15:40 | comment | added | David E Speyer | @NM Yup, exactly. And I think that your more general conjecture is the same thing, applying an element of $SL_2(\mathbb{R})$ (which is a hyperbolic symmetry, but changes the Euclidean area). It's weird, I've never seen the Euclidean areas of either Klein or Poincare triangles come up. | |
Sep 19 at 15:26 | comment | added | N M | Here is another interpretation of this answer. If we view $x/y \in \mathbb{RP}^1$ as an ideal point in the Poincaré half-plane model, the point $\phi(x,y)$ is the corresponding ideal point in the Klein disk model. Hence the triangulation in this answer is simply the Farey triangulation in the Klein disk model. | |
Sep 18 at 19:31 | comment | added | Nikita Kalinin | very nice, thank you! I will post my solution tomorrow. My solution is simpler, but $\pi$ appears there as the angle. And you've got $\pi$ as the area! That is fantastic! | |
Sep 18 at 19:14 | comment | added | David E Speyer | The odd thing is that the idea of putting triangles inside a circle works, but the way to put the circle into the triangle is by the "Pythagorean map", not by rescaling! | |
Sep 18 at 19:13 | comment | added | David E Speyer | In our problem, we have each triangle $(0, \vec{x}, \vec{x}+\vec{y})$ has area $1$, and we sum up $\tfrac{1}{|x|^2 |y|^2 |x+y|^2}$, so we are summing up the fraction that each of these circles occupies in its circumcircle. So that made me think of rescaling the triangles to all fit in the same circle, and adding up the areas. | |
Sep 18 at 19:13 | comment | added | David E Speyer | I came at this in an odd way. If a triangle has side lengths $a$, $b$, $c$, circumradius $R$ and area $K$, then $K = \tfrac{abc}{4R}$. So $\tfrac{K}{\pi R^2} = \tfrac{16 K^3}{\pi a^2 b^2 c^2}$. In other words, $\tfrac{16 K^3}{\pi a^2 b^2 c^2}$ is the fraction of the circumcircle occupied by the triangle. | |
Sep 18 at 16:56 | history | answered | David E Speyer | CC BY-SA 4.0 |