Same as you did. $F*(\Vec_\Pt)=\lim_{x\to\infty}\frac{d}{dt}F(\Pt+t\Vec)$ — DOTDO Oct 10 '16 at 1:42

Same as you did. $F*(\Vec_\Pt)=\lim_{x\to\infty}\frac{d}{dt}F(\Pt+t\Vec)$ — DOTDO Oct 10 '16 at 1:42

In the picture, $\Vec{v}$ is an arbitrary element of $\mathbf{R}^{3}$; the vector $(1, 0, 2)$ lies on the line $E_{-1}$ at an unspecified position, and $E_{1}$ is the plane orthogonal to $(1, 0, 2)$. The points are: (i) Every vector in the plane is fixed by the reflection $L$; (ii) Every vector on the line $E_{-1}$, i.e., every scalar multiple of $(1, 0, 2)$, is mapped by $L$ to its negative; (iii) The action of $L$ on an arbitrary vector $\Vec{v}$ may be seen by decomposing $\Vec{v}$ into eigenvectors of $L$ and letting $L$ act on the eigenvectors by scalar multiplication. — Andrew D. Hwang Apr 23 '17 at 16:16

I don't know that this criterion has a common name. If you "dualize" with respect to the Euclidean metric, the vector field $\Vec{p}$ becomes a differential $1$-form; ( * ) says the form is closed, and on a simply-connected domain every closed form is exact (see de Rham cohomology). In a general domain, (*) plus "the line integral vanishes around a set of loops forming a basis of $1$-dimensional homology" is equivalent to being a gradient field. — Andrew D. Hwang Mar 15 '17 at 21:38

@mfripp: The display above the equation you ask about gives an expression for $\frac{\dd p_{i}}{\dd x_{j}}$. Swap $i$ and $j$ to get $\frac{\dd p_{j}}{\dd x_{i}}$, equate using (*), and cancel the terms that are obviously invariant under exchanging $i$ and $j$: the second partial, and the product of the two first partials divided by $(\Del u \cdot \Vec{x})^{2}$. — Andrew D. Hwang Mar 14 '17 at 19:12

Depending on $d$, $\Vec{x}$ could be something of a mess; that's why I'd pick an orthonormal basis for the hyperplane $H$ orthogonal to $d$, then use that basis to define a Cartesian coordinate system in $H$. Finding that basis may be a pain, but it's algorithmic using Gram-Schmidt. (Not sure I understand the components of the $d$ you mention; my browser's math fonts may be a bit wonky, but it looks like $d_{1} = d_{1 + (n+1)} = d_{1 + 2(n+1)} = \dots = d_{n} = -1/\sqrt{n}$...?) — Andrew D. Hwang Jan 2 '17 at 15:43

Do you mean you want to know in detail how to create a parametric curve to trace a conic in a particular plane? (The short answer is: Each of $\Vec{x}_0$, $\Basis_1$, $\Basis_2$ is an ordered triple of real numbers, so the given formula for $\Vec{x}(t)$ can be expanded to an ordered triple of functions of $t$.) — Andrew D. Hwang Oct 18 '16 at 13:49

That's a good strategy, but I actually think I still struggle when following these steps. It's the second part of phase 1 that gives me trouble. How do I work backward? Sometimes it's not too clear what I can do from the inequality $|f(\Vec{x}) - L| < \eps$ — Matt24 Sep 25 '16 at 15:20

@goblin: Actually, I'd prefer to denote the components $x^{j}$ (in the spirit of covariant and contravariant indices) and then write $[\Vec{x}]^{S}$ to indicate "$S$ comprising all the indices". :) Your multiplicative notation still works as a formal dot product of the coordinate vector with the basis denoted as a column having basis elements as entries:$$\Vec{x} = [\Vec{x}]^{S} \cdot S = [\Vec{x}^{j}] \cdot [\Vec{v}_{j}] = \sum_{j} x^{j}\Vec{v}_{j}.$$ — Andrew D. Hwang Jun 4 '16 at 15:59