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8m
comment Seperating points in the complex plane
@MarkMcClure: By the time I'd finished typing, you'd posted your comment. Wikifying was a compromise between deleting my answer and stealing yours. :)
13m
answered Seperating points in the complex plane
15h
answered critical point for the curvature does not correspond to a local maximum/minimum.
19h
comment What is a transformation that can't have shearing called?
@WillJagy: Entirely agreed if the OP did mean "preserve arbitrary rectangles", though the wording "cannot have shearing or scaling after rotation" still suggests to me the OP meant "axis-aligned". :)
20h
comment Where is the “thread” of a river?
The medial axis of the region enclosed by the banks might be a reasonable definition. (This seems to be what @Meelo is getting at.)
20h
comment What is a transformation that can't have shearing called?
Since the set isn't closed under composition (as you say, you can't do unequal scaling after rotation; your claim about "rectangles mapping to rectangles" is true only for axis-aligned rectangles), it's conceivable these mappings have never attracted "serious" attention. Anyway, I've never seen a name given to them, but if it matters, you're looking at affine mappings whose linear part is a $2\times 2$ real matrix of the form $OD$, with $O$ orthogonal and $D$ (invertible and) diagonal.
20h
answered Question about series convergence $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$
1d
comment Gauß and mean curvature
If you're seeking a computational argument, it may help to notice that $N \circ f^{-1} = (N \circ \Phi) \circ (f \circ \Phi)^{-1}$ for every reparametrization $\Phi$.
1d
comment Gauß and mean curvature
Ah...! What you call "the Gauss map" I'd call "the Gauss map in local coordinates". To me, $N \circ f^{-1}:M \to S^{2}$ is "the Gauss map". Anyway, does it make sense that $N \circ f^{-1}$ maps a point of $M$ to "the" unit normal at that point, and that $-D(N \circ f^{-1})$ has the stated geometric interpretation/meaning?
1d
comment Gauß and mean curvature
(Just noticed I've used "$M$" to denote the surface. I'd avoid "$S$" for the surface, lest we use a single letter to denote a surface, its shape operator, and possibly also the unit sphere.) I don't understand your notation "$L = -DNDf^{-1}$": If $f$ is a parametrization of $M$, and if $N:M \to S^{2}$ is the Gauss map, then $DN$ and $Df^{-1}$ aren't composable (regardless of whether "$Df^{-1}$" means $(Df)^{-1}$ or $D(f^{-1})$...? In case it helps, O'Neill's definition of the shape operator is (essentially) $-DN$.
1d
comment Gauß and mean curvature
The value $S_{p}(v)$ is the (signed) curvature at $p$ of the plane section of $M$ by the plane spanned by $v$ and the unit normal $U$, i.e., the reciprocal radius (up to sign) of the osculating circle to this plane section. The Wikipedia pages for the second fundamental form and principal curvature may be helpful.
1d
comment Why Do The Axioms of Euclidean Geometry Not Need To Include the Definition of Space?
@timax: It seems to me your primary objection/unease (and usage of the verb "exist") is philosophical, not mathematical. Separately, your post poses multiple questions covering a wide range. It's possible this site's format isn't a good match, unless you can focus your question...? Be that as it may: 1. Yes, $n$-dimensional Euclidean geometry has been studied; 2. It's (mathematically) possible to have two points without having a one-dimensional "ambient universe".
2d
answered Gauß and mean curvature
Jan
21
comment Compute a parallel transport
@Odile: You're welcome. :) It's crucial to this argument that on a sphere, geodesics lie in a plane. Think of a helix on a circular cylinder in $\Reals^{3}$.
Jan
21
answered Compute a parallel transport
Jan
20
comment Gradient of a parametric form
@uitty400: Your parametric equation for a plane is fine; the variables are $s$ and $t$. :) (Or are you counting the surface parameters $(u_{0}, v_{0})$ as well? Those are fixed to describe a tangent plane at one point. This issue is related to my uncertainty about what exactly you want to plot.)
Jan
18
comment Why is $|x|$ not differentiable at $x=0$?
@MathN00b: That's not the usual definition. :) Here's why it's not a "good" definition: If your graph were a horizontal line, say $y = 0$, then every non-horizontal line would be "tangent" by the "touches at one point" criterion. If instead your curve were a parabola, say $y = x^{2}$, then every line whatsoever would be tangent according to "touches at one point (in a neighborhood)".
Jan
18
comment Why is $|x|$ not differentiable at $x=0$?
@MathN00b: What is your definition of a "tangent line"? :)
Jan
18
answered If $f\circ g = g \circ f$ does that mean that both functions are to and from the same set and both are bijections? Does it tell us anything else?
Jan
18
comment Let P be a polynomial in R^2 and $\oint$ P(x,y)(dx + idy) = 0 for every circle $\gamma$ in $\mathbb{C}$. Prove that P is $\mathbb{C}$-differentiable.
Just asking: Do you know (of) the Cauchy-Riemann equations (and their significance)?