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14h
comment Imaginary curvature.
If you're coming from a background of popular expositions, it's going to be all but impossible to ask (!) or answer your question at the depth you seek: Merely defining "classical" notions of curvature entails either calculus or substantial hand-waving. Still, it may help to skim the obligatory Wikipedia pages on curvature and Gaussian curvature if you haven't already.
18h
answered Geometric interpretation of an integral inequality
Aug
19
reviewed Close How to find this integral $\int_{0}^{\infty}\dfrac{f(x)}{g(x)}dx$
Aug
19
reviewed Close Example of a function $u\in L^\infty(0,T,H^1)$ such that $u_t\notin L^\infty(0,T,H^1)$
Aug
19
answered $x\cdot 0 \neq 0$ infinitely many zeroes on a finite interval
Aug
18
comment Surjectivity of an integration map
Uncannily, I was typing up precisely the same counterexample.... :)
Aug
18
revised Interpret to a complex plane!
Removed "complex-geometry" and "calculus" tags, added LaTeX, tweaked wording, punctuation.
Aug
17
comment Manifolds, charts and coordinates
If it matters, $\phi$ and $\theta$ aren't coordinates (in any reasonable sense) at the poles. Geographically, longitude fails to be well-defined at the poles in a more degenerate way than longitude fails to be well-defined elsewhere on the sphere. It does turn out that for every compact manifold $M$, there exists a coordinate chart covering a dense open subset of $M$. However, I don't know of a "low-tech" proof, e.g., one using only basic definitions.
Aug
17
comment Manifolds, charts and coordinates
Could you explain more about how you've seen one set of coordinates to cover the $2$-sphere? There's no obvious analogy with your example of using an angle parameter $\theta$ on the circle.
Aug
17
answered Manifolds, charts and coordinates
Aug
17
comment Series of polynomials and uniformly convergence
The Weierstrass test looks fine: The right-hand sum (in your preceding comment) is bounded by $\exp\bigl(\max\{|z|, 1\}\bigr)$, and $|z|$ is bounded on compact sets.
Aug
17
reviewed Close Equation of parabola with focus and tangent
Aug
16
comment Seeking graphics of elliptic curves as surfaces
FWIW, here is (half of) a smooth cubic, and here is a nodal cubic, with the real points highlighted on each.
Aug
13
comment The Kähler form and the anticanonical line bundle
Something looks strange with your stated criterion, in that neither $K(M)$ (homology classes of curves in $M$) nor the Kähler form depend on the anticanonical class. This may be a long shot, but perhaps you're looking for Kleiman's criterion, and the intended condition is something like "the anticanonical class is positive on $K(M)$"...?
Aug
13
comment Equation of a straight line in spherical coordinates
Perhaps the lunes argument obviates your question, but in case it's helpful: A great circle on the unit sphere $S$ is the intersection of $S$ and a plane $P$ through the origin. To get an equation in spherical coordinates, pick a unit vector $(a, b, c)$ normal to $P$, and write $ax + by + cz = 0$, replacing $(x, y, z)$ by their spherical coordinates expressions. (There are multiple conventions; not sure which you're using.) The result is a linear relation between circular trig functions of angles (rather than between angles themselves) and probably not pleasant for your purposes.
Aug
13
comment Equation of a straight line in spherical coordinates
So that's "yes". :) Is there a reason you want a proof other than the standard geometric argument using lunes?
Aug
12
comment Equation of a straight line in spherical coordinates
By "angle sum formula", do you mean something equivalent to the theorem that the sum of the interior angles (of a spherical triangle) minus $\pi$ is the area of the triangle divided by the radius (of the sphere) squared?
Aug
11
comment Prove that $f(a) \leq f(x) \leq f(b) $
This argument (correctly) shows $f$ is "increasing at $x$", but not that $f$ is increasing on some neighborhood of $x$. The latter is generally false without an additional hypothesis, e.g., $f'$ is continuous (and positive) at $x$, or (weaker) that $f'$ is strictly positive in a neighborhood of $x$.
Aug
11
comment Why do objects that are farther away look smaller?
Small coda on this (excellent) answer: The small angle approximation presumes space is (approximately) flat. In a positively-curved space it's generally untrue that apparent angular size decreases with distance. (!)
Aug
10
answered Length of Difference Curve