froggie
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 Apr 20 comment Component test for conservative fields The curl is a vector with three components; if the curl is 0, each of its three components must be 0. The 3 equations you wrote end up being exactly the equations saying that all three components of the curl are 0. I guess the point is, once you know about the curl, the conditions of the component test are easy to remember, since they can be rewritten simply as "curl F = 0" (which is pretty easy). Until you've learned about the curl though, I agree the component test looks really random. Apr 20 comment Component test for conservative fields Are you familiar with the curl of a vector field? The conditions of the "component test" are equivalent to the curl of the vector field being 0. The curl is an object which encodes the rotational behavior of the vector field; non-zero curl implies that the vector field has non-zero circulation "around the curl" Apr 7 answered Rolle's Theorem for Complex Functions Apr 3 revised The existence of the roots of an holomorphic function on an open connected domain Fixed some notation. Still a massive run-on... Jan 26 awarded Yearling Dec 14 awarded Revival Nov 19 answered Convergence of measures and potential theory Nov 19 comment Covering map of the annulus What have you tried? Nov 18 revised Which functions have a list for all periodic points of them? added 30 characters in body Nov 18 answered Which functions have a list for all periodic points of them? Nov 18 comment Integral under Diffeomorphsim, sign of Jacobian That is another way to say it, yes! Nov 18 answered Integral under Diffeomorphsim, sign of Jacobian Nov 15 comment Equation for plane perpendicular to curve tangent hint Yes, that is correct. Nov 15 comment Lebesgue measure of proper subset What happens to the measure of a closed interval if you remove one of its endpoints? Nov 15 answered Equation for plane perpendicular to curve tangent hint Nov 12 answered Gradient of a function twice Nov 10 comment Maximization via Lagrange multipliers vs. substitution and partial derivatives I'm not advising one over the other. However, in your comment, there is what sounds like a mathematical error: one must consider the possibility which maximizing a function on $z^2 + y^2\leq 5$ that that maximum lies on the boundary $z^2 + y^2 = 5$. If that were the case (it's not for this particular example but could be in another example) the maximum may not appear as a local maximum of the unconstrained function. Nov 9 comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$? I think the idea of "outside" can really be subtle. For instance, if your surface is not orientable (like a Mobius band), then the word "outside" really has no meaning. But even for something like the triangle in your pictures: why is the normal vector you've drawn the "outside" one? Or imagine two cubes, one whose top face is a square in the $xy$-plane and another of the same size sitting on top of the first, so that its bottom face is the same square in the $xy$-plane. For these cubes, the outer normal vectors will be pointing different directions on this square. Nov 9 comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$? Assuming it is orientable (so for instance, not a Mobius band), a surface in $\mathbb{R}^3$ always has a continuously varying (even smoothly varying) unit normal vector field. As for the notion of "outside" this only makes sense if the surface you're considering is closed. For instance, the sphere is closed, so it makes sense to talk about an outward unit normal, but the triangle in your pictures is not closed, so it is a choice to call the arrow you've drawn "outward" while the opposite is "inward". Nov 9 answered If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?