My first instinct is to average the vertices. My second instinct is to go, "Wait, didn't I already establish that that gives the wrong answer a few months ago?"
Well, it doesn't matter if the origin is inside, because we can just create an "artificial origin" that's inside the shape, so I don't think that's any problem
We know that the centroid breaks up into equal areas. We also know that the medians are broken into a ratio of 2:3. Thus, we can use this to create some form of a system of equations with 1/2*b*h.
'Cause we can replace each triangle with a point mass without changing the location of the centroid, and then the centroid of two point masses lies on the line segment between the two points
But in any event, it was incomprehensible, the determinant just happened out of nowhere, and at the time we were black boxing Lebesgue differentiation even though we never knew the Lebesgue integral
If you scale a single vector, you see that the area scales, and then to add just do it to each separately. Then it's alternating because if you duplicate a vector, stuff lives in a lower dimensional space. And it sends the identity to 1 because unit cube
Given two vectors, $a$ and $b$, you have that the map $f_{a,b}(v) = \det(v,a,b)$. This is a linear functional on $\mathbb{R}^3$, so by Riesz you can find some vector $a\times b$ such that $f_{a,b}(v) = \langle v, a\times b\rangle$
And then the properties of the cross product are clear as day. $a\times a = 0$, it gives a vector that's orthogonal to $a$ and $b$ since you just let $v = a$ and $v = b$, its norm is by the area of determinant business, etc
@0celo7, I just learned that $f:\Bbb C\to\Bbb C $defined by $f(z)=|z|^2$ is differentiable at $z=0$ only, but if $z=x+iy$ then Frechet derivative of $f(z)=x^2+y^2$ exist at all points in $\Bbb C$. I was wondering from which directions does difference quotient approach in this Frechet derivative.
Doesn't the Fréchet derivative just mean that the function $\Bbb R^2\to\Bbb R^2$ defined by $\begin{bmatrix}x\\y\end{bmatrix}\mapsto \begin{bmatrix}x^2+y^2\\0\end{bmatrix}$ is differentiable as a function on a real vector space?
Unless I misunderstand what a Fréchet derivative is (very possible)
@0celo7 i know that, but the solution that says $f(z)$ complex differentiable computes difference quotient for both real axis and imaginary axis, and then derives that the limit of difference quotient can be equal for real axis difference quotient and imaginary axis difference quotient at 0 only. Hence i was wondering what direction frechet derivative ttakes limits.
I want to find a norm on $\Bbb R^2$ in which the unit circle has a hexagonal shape with the edge at $(1,0)$. I found this question math.stackexchange.com/questions/1594480/hexagon-boundary but why does it say max instead of min?
@philmcole a line is of the form ${\bf n}\cdot (x,y)=h$, where $\bf n$ is a unit normal vector and $h$ is the distance from the line to the origin. the line segment at the top of the hexagon would use ${\bf n}=(0,1)$, so the dot product would just be $y$, and the line segment one sector clockwise would have angle $\pi/6$, so correspond to ${\bf n}=(\frac{\sqrt{3}}{2},\frac{1}{2})$. and so on the the rest.
think about the dot product ${\bf n}\cdot(x,y)$ in those cases. pairing antipodal ones together, we get absolute values, and then thinking algebraically we get the $\sqrt{3}|x|+|y|$ one.
Can anyone help me in this regard : The equation of directrix of an ellipse is $x+y-1=0$ and the focus near to it is $(-1,1)$ . The eccentricity is $\frac{1}{2}$ . How can I find the vertex of ellipse ?
@philmcole let $\bf p$ be the point on the line closest to the origin, at a distance of $h$. the unit normal vector $\bf n$ points straight to the point $\bf p$ then. so ${\bf n}\cdot{\bf p}=h$. if $\bf x$ is any other point on the line, then the displacement ${\bf x}-{\bf p}$ is perpendicular to $\bf n$, so ${\bf n}\cdot({\bf x}-{\bf p})=0$, which rearranges to ${\bf n}\cdot{\bf x}=h$
@anon Thank you! I got it. I wonder why we've never learned this in class. We only learned the explicit representation as $\mathbf{x} = \mathbf{s} + t \mathbf{r}$ with $\mathbf{s}$ being the support vector and $\mathbf{r}$ the direction vector. Though for planes we learned the explicit and implicit representation...
Hi, how can I use the Banach fixed point theorem to show that a certain recursive sequence converges? Suppose I have a sequence $(x_n)_n$ which is defined recursively by $x_{n+1}=T(x_n)$ and I know that $T$ is a contraction. By the BFP theorem I know there exists a fixed point $x_0$ s.t. $x_0 = T(x_0)$. How can I use that to show that the sequence converges?
Regardless of however one interprets this tweet, I think he needs a good dose of jab for making sloppy commentary on a complex socio-political historical era just for publicity.
If $U,A,B$ are subgroups of a Lie group $G$ such that $U$ is dense and $AuB$ is dense for all $u \in U$. Can I choose for all $g \in G$ sequences $(a_n),(b_n),(u_n)$ with $a_nu_nb_n \to g$ and $u_n \to e$?
I tried, starting with $u_n$, but I can't get the sequences $a_n$ and $b_n$.
Given a surface $S$ in a contact $(2n+1)$-manifold $(M,\xi)$ one can define the characteristic foliation of $S$ via looking at where the tangent bundle to $S$ coincides with $\xi$ (the singular part of the foliation), and otherwise where it intersects.
It turns out that locally, if $\theta$ is a...
This was a question I asked some time ago and I feel like it should be answered easily with just general knowledge of differential geometry but I don't know.
It's in the context of contact geometry, but it doesn't seem to be appealing to anything in contact geometry.
