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I hereby authorize whoever it may concern to post my comments as answers if they wish, whether as community wiki or otherwise if you like imaginary internet points.


1d
comment Equation for Circle in 3D Space Given Center, Radius, and Point
Knowing the center and just one point does not fix the circle. For example, if the center is at the origin and the point is on the $x$-axis, you don't know if the circle lies on the $xy$-plane, or the $xz$-plane, or anywhere in between. If you knew two points then you could get somewhere.
1d
comment The unit vector in the direction of u
It's hard to comprehend because it's false. Let $u=(1,1,1)$. Then $u\cdot u=3$ and $\|u\|^2=(\sqrt3)^2=3$, so $u\cdot u=\|u\|^2$ and the statement is true. Again, I have no idea where you are getting the $1$ from.
1d
comment The unit vector in the direction of u
The dot product of $(1,1,1)$ with itself is not $1$, it's $3$. I have no idea which $1$ you're talking about in your answer.
1d
comment The unit vector in the direction of u
The norm of $(1,1,1)$ is not $1$, it's $\sqrt3$.
2d
comment Is Frobenius norm induced up to a scalar factor?
I didn't expect this answer to turn into "Please prove as many properties of induced norms as possible", but here you go: From the definition of induced norm, one can prove submultiplicativity as follows: $\|AB\| = \max\dfrac{\|ABx\|}{\|x\|} \le \max\dfrac{\|A\|\|Bx\|}{\|x\|} = \|A\|\max\dfrac{\|Bx\|}{\|x\|} = \|A\|\|B\|$. First equality is the definition of $\|AB\|$, inequality is because $\|A\|\ge\dfrac{\|Ay\|}{\|y\|}$ for all $y$ and so $\|Ay\|\le\|A\|\|y\|$, last equality follows from the definition of $\|B\|$.
2d
comment Is Frobenius norm induced up to a scalar factor?
@user153012: Sure, your argument shows that the proposed norm is not submultiplicative, so it cannot be an induced norm. However if you want to argue that it is also not a matrix norm at all, you should be aware that not all authors require matrix norms to be submultiplicative.
2d
comment Is Frobenius norm induced up to a scalar factor?
@user153012: Edited. You can also prove that the induced norm of an identity matrix is $1$ in a similar way, simply by plugging in the definition.
Aug
26
comment Geometry question pertaining to a plane going through the skeleton of a cube
$|BF|+|FE| = \big(|BC|+|CF|\big)+\big(|FD|+|DE|\big) = |BC|+\big(|CF|+|FD|\big)+|DE| \ge |BC|+|CD|+|DE|$
Aug
25
comment Is a Squircle a circle?
"This was posted as an answer, but it does not attempt to answer the question"...
Aug
24
comment How to apply Gaussian kernel to smooth density of points on 2D (algorithmically)
Cross-posted to the stats site: stats.stackexchange.com/q/113078
Aug
24
comment Ellipse like on sphere
Don't use TinyPic. Just upload the image using the "Image" button in the toolbar (sixth from the left).
Aug
23
comment Is there an easier way to find the “natural” integration constant?
Don't trust everything Mathematica tells you. Take $f(x)=x$, and the inner integral becomes $\int_{-\infty}^{+\infty} t e^{i\omega t}\,\mathrm dt$ which doesn't converge for any $\omega$.
Aug
23
comment Is there an easier way to find the “natural” integration constant?
I don't understand. Doesn't the inner integral diverge for any polynomial $f$?
Aug
22
comment Weighted least squares with angular data
In that case, perhaps you should mention somewhere in the question that $z_{ij}=\theta_i-\theta_j$ is only true modulo $2\pi$.
Aug
22
comment Weighted least squares with angular data
Copying my comment on a deleted answer: This doesn't account for the angular nature of the data. For example you could have $z_{12}=2π/3$, $z_{23}=2π/3$, $z_{34}=2π/3$, $z_{41}=0$ which can be satisfied exactly but not with your method.
Aug
22
comment Is there an easier way to find the “natural” integration constant?
It seems to me that your series converges at $s=-1$ extremely rarely: it diverges whenever $f$ is a polynomial, for example. Perhaps extrapolation via polynomial interpolation formulas is not the right tool for the job.
Aug
21
comment Why is that *any* union of open sets is open but only *finitely many* intersections of open sets is open?
Previously: Why can the intersection of infinite open sets be closed?
Aug
21
comment Why is that *any* union of open sets is open but only *finitely many* intersections of open sets is open?
If you allow the intersection of infinitely many open sets to be open, then you have to accept, say, $\{x\}=\bigcup_{n=1}^\infty \bigl(x-\frac1n, x+\frac1n\bigr)$ as open for any $x\in\mathbb R$, and from there it's a hop, skip, and jump to having every possible subset of $\mathbb R$ be open.
Aug
21
comment From 2 to 3 dimensions: integrating a force along a contour/surface.
In that case, the integral becomes $\oint(\mathbf x-\mathbf q)\times\mathrm d\mathbf x$ where the integral is taken along the contour of tangential points. One can see that the integral is independent of $\mathbf q$ because $\oint\mathrm d\mathbf x=0$, so you might as well take it to be the origin, yielding $\oint\mathbf x\times\mathrm d\mathbf x$.
Aug
17
comment Angular momentum of an accretion disk
The angular momentum of a system can change if momentum enters or leaves the system, for example if new material falls into the accretion disk from outside, or material in the disk disappears into the black hole.