Reputation
18,520
Top tag
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
3 15 45
Newest
 Yearling
Impact
~147k people reached

1d
comment Branch cut of $\sqrt{z}$
With the proposed half-open interval, the square root function is defined on the whole plane, but is discontinuous along the negative real axis. By taking the domain to be the slit plane, the square root is holomorphic throughout its domain.
2d
comment Finding the maximum of a function on $ \Bbb{S}^{7} $.
Perhaps not terribly important, but your question title mentions $S^{8}$, while your constraint is the unit sphere in $\mathbf{R}^{8}$, i.e., $S^{7}$. :)
Aug
28
answered Geometric interpretation of inverse complex function?
Aug
26
revised Questions about algebraic curve definition
Substantially expanded to address issues raised in the comments
Aug
26
comment Questions about algebraic curve definition
@bsdshell: To clarify, are you asking 1. Why does a homogeneous polynomial contain only monomials of the some fixed degree? 2. Why must we use homogeneous polynomials to cut out a plane curve in projective space? or 3. Something else?
Aug
26
answered Questions about algebraic curve definition
Aug
25
revised how to find the locus when distance from the origin is defined as d(x,y) = max { |x|,|y|},d(x,y) =a (where 'a ' is a non zero constant )
edited tags
Aug
25
comment Kernel of the linear operator from $P_{5}$ to $R^{2}$
Small linguistic point: When you speak of "the basis", you're implicitly asserting that a basis is unique, which (except as noted below) is never the case. Instead, speak of "a basis". :) (Tangential exercise: Characterize the vector spaces having a unique basis. Up to isomorphism, precisely one has positive dimension.)
Aug
25
answered Prove that $\sqrt{3}$ is not a rational number
Aug
24
answered What hyperbolic space *really* looks like
Aug
23
comment Plotting a skew ellipse
To do spherical geometry, it seems you'd want to represent and compute with great circles (and points) in three dimensions, then project them to the plane, rather than plotting directly to the plane...? In any case, Patrick Ryan's Euclidean and Non-Euclidean Geometry may be worth a look.
Aug
22
comment Manifold is not orientable
@Fallen: Yes, the converse is true. Intuitively, if $M$ is non-orientable, there exists a loop (closed curve) around which "the orientation reverses". Thicken the loop into a tube and cover it with two "solid cylinders" $U_{a}$ and $U_{b}$. There are details to check, and this sketch may not be the easiest strategy to make rigorous. (E.g., if no such charts exist, start with an arbitrary atlas and reverse chart orientations as needed to get an atlas of compatible charts. Again, though, there are details to check....) Perhaps post a separate question?
Aug
22
answered vertical asymptote - is it possible to have one like this?
Aug
21
answered Manifold is not orientable
Aug
21
comment Looking for a biholomorphism
If you want "conformal" you'll have to specify metrics on the domain and target.... Are you willing to take a hyperbolic plane crossed with a Euclidean line as your half-space of $\mathbf{R}^{3}$?
Aug
21
comment Manifold is not orientable
If you pick a maximal orienting atlas for $M$, then arbitrary charts $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ are either compatible or "anti-compatible" with the chosen orientation. If both charts are compatible, the transition function $\phi_{ab}$ is a transition function for an orienting atlas; if either chart is anti-compatible, post-composing with a coordinate reflection defines a compatible chart. (I'd personally structure the proof as a contrapositive, not contradiction: Assume $M$ is orientable, show an arbitrary transition function has Jacobian determinant of constant sign, done.)
Aug
20
comment How is this circle inversion formula calculated?
You're very welcome. :)
Aug
20
comment How do 3 points define a plane?
Since this has gotten bumped: Two distinct points determine a unique line in space; a third point not on this line determines a unique plane. If you watch dust motes drifting in a shaft of sun, the claim will quickly become intuitively obvious. :)
Aug
20
answered How is this circle inversion formula calculated?
Aug
20
comment How is this circle inversion formula calculated?
Do you know the image of a circle under inversion is itself a circle? If so, that simplifies the proof.... :)