Activity for Derek Elkinsâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #286848 |
This doesn't really make a lot of sense currently. It's also doesn't seem to be about graphs. You don't seem to care about any of the connectivity information, while you do seem to care about geometric information that is usually not considered part of a graph. For example, images $G_2$ and $G_3$ wou... (more) |
— | over 2 years ago |
| Edit | Post #286833 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286833 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Show that $f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2}$ $\infty$ is not a number so $\arctan(\infty)$ isn't meaningful, nor does it have a reasonably standard interpretation. You can already tell something is wrong, because there's nothing in your argument that keeps it from applying for all values of $x$. One way to attempt to make an argument along the ... (more) |
— | over 2 years ago |
| Edit | Post #286174 |
Post edited: fix typos |
— | over 2 years ago |
| Comment | Post #286034 |
If you want to talk about square roots of -1, you need to define what "-1" means and what multiplication means so that you can solve the equation $x \cdot x=-1$. If you want to use the embedding $(x,y)\mapsto(x,y,0)$, that's fine, but you haven't defined a multiplication on triplets nor what -1 means... (more) |
— | over 2 years ago |
| Edit | Post #286482 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Are vectrices useful for calculations as opposed to formalism? This is largely just a notational difference insofar as it's used in the example you give. While it often is convenient and enlightening to view a matrix as a "vector" of vectors momentarily, taking this too seriously can be problematic. A vector space expects a field) of scalars, and vectors don't f... (more) |
— | over 2 years ago |
| Edit | Post #286174 |
Post edited: Add some additional historical context |
— | over 2 years ago |
| Comment | Post #286034 |
Your constraints amount to asking for the quaternions whose plane of rotation is represented by a bivector with $0$ $y$-$z$ component. This will restrict to a 2D subspace of bivectors, but there's nothing special about this. The space of rotation operators with $0$ $y$-$z$ component, however, is not ... (more) |
— | over 2 years ago |
| Comment | Post #286034 |
In quaternion algebra, $1$ is not identified with the vector $(1,0,0)$. If we used quaternion terminology for complex numbers, we'd say complex numbers have a scalar part (the real part) and a 1D vector part (the imaginary part). These spaces are (double covers of) spaces of rotation operators that *... (more) |
— | over 2 years ago |
| Comment | Post #286034 |
That part is talking about orthogonal planes. But where are you getting triplets in 2D? For the complex plane, we usually identify $(1,0)$ with 1 and $(0,1)$ with $i$. With complex multiplication, we then get $i^2 = -1$ and, modulo sign (the negation is not orthogonal), that's the only square root of... (more) |
— | over 2 years ago |
| Edit | Post #286174 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Complex numbers in 2D, quaternions in 4D, why nothing in 3D? There is a famous result) about finite-dimensional real associative division algebras. You may find that interesting, but I don't think it's what you want. It is a direct answer to the question of why we "can't" multiply "triplets of numbers" if you require that multiplication to satisfy certain prop... (more) |
— | over 2 years ago |
| Comment | Post #286034 |
Even in 3D, the notion of axis is problematic. Physicists have a distinction between the notion of an "axial vector" and a "position vector". Reflecting a "position vector" in a plane orthogonal to it negates it, while reflecting an "axial vector" does nothing. When stated as a fact lacking context a... (more) |
— | over 2 years ago |
| Comment | Post #286034 |
Yes, I was implying orthogonal when I said it produces 3 as I said explicitly earlier in the comment. Even if you don't limit to orthogonal, infinitely many is still more than 1. For your latter question, in 4D with an orthogonal basis labelled $x$, $y$, $z$, $w$, what is the "axis" orthogonal to the... (more) |
— | over 2 years ago |
| Comment | Post #286034 |
This answer started off good but stumbled pretty badly near the end. Instead of talking about rotations about an axis, talk about rotations in a plane. In 2D, which we can identify with the complex plane, there is obviously one plane in which to rotate. Going up a dimension to 3D doesn't give us just... (more) |
— | over 2 years ago |
| Comment | Post #285978 |
The same way you'd convert any linear function between finite dimensional spaces to a matrix. Choose a basis for the input, i.e. $\vec x$, and a basis for the output, which will be a basis of bivectors, and evaluate the linear function for each basis vector and expand the result in the output basis.... (more) |
— | over 2 years ago |
| Comment | Post #285978 |
Say $\vec y = \mathbf e_1$ in a set of orthogonal basis vectors $\\{\mathbf e_i\\}_{i=1}^m$. While there are $m \choose 2$ basis bivectors, the only ones we care about are of the form $\mathbf e_1 \wedge \mathbf e_i$ for $i \neq 1$ (as $\mathbf e_1 \wedge \mathbf e_1 = 0$), so there are $m-1$ constra... (more) |
— | almost 3 years ago |
| Edit | Post #285978 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: Maximize Independent Variable of Matrix Multiplication Since all the $xi$ are restricted to be non-negative, the magnitude of the vector increases if and only if the sum of the $xi$ increases. Therefore, maximizing $\sumi xi$ is equivalent to maximizing $|x|$. This sum is a linear function. Your direction constraint can be expressed as: $$\vec y \wedg... (more) |
— | almost 3 years ago |
| Edit | Post #285442 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: Are we in a "history-valley" for Topology? I'm going to focus on the "history-valley" aspect, and to jump to the upshot: No, I don't think we're in a "history-valley" with respect to topology. I more or less agree with Guilherme Gondin†that the significant thing in the ??? period was the foundational "crisis" (spurred on by the kind of is... (more) |
— | almost 3 years ago |
| Comment | Post #285436 |
Topologies formulated as sets of open subsets are closed under (arbitrary) unions and *finite* intersections. Requiring countable intersections of open subsets to be open would break many standard examples of topologies. You also do need to explicitly include the total set as well. (If you were talki... (more) |
— | almost 3 years ago |
| Edit | Post #285424 |
Post edited: Fix braces. |
— | almost 3 years ago |
| Suggested Edit | Post #285424 |
Suggested edit: Fix braces. (more) |
helpful | almost 3 years ago |
| Comment | Post #285365 |
As has been mentioned to you before, you should inline relevant text rather than or in addition to using an image. Ideally, you should inline it so that there is no benefit to an image, but if you are really concerned about transcription errors or are unable to reproduce some notation, then including... (more) |
— | almost 3 years ago |
| Comment | Post #285371 |
As it states and you can derive, $B > C$ is logically equivalent to $C = 2$ in this scenario. Thus the expressions $P(C > D \mid C = 2)$ and $P(C > D \mid C \neq 2)$ are the same as $P(C > D \mid B > C)$ and $P(C > D \mid B \not> C)$ respectively. If $C > D$ was independent of $B > C$, then these exp... (more) |
— | almost 3 years ago |
| Edit | Post #285384 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: Acceptable, usual to write $\ge 2$ pipes simultaneously? As mentioned in the answers you referenced in earlier versions of your question, $(-\mid-)$ is not standalone notation in usual probability theory notation.^[I have seen it used, e.g. in Jaynes' "Probability Theory: The Logic of Science", as standalone notation, but not in a way such that $P(A\mid B)... (more) |
— | almost 3 years ago |
| Comment | Post #284788 |
@#53922 Calling a question a "soft question" doesn't suddenly make it being overly broad, primarily opinion based, and/or lacking context or research okay. (more) |
— | about 3 years ago |
| Comment | Post #284718 |
Finally, there is some context which isn't *technically* required but omitting it sets up confusion down the line. First, most discussions of tensors by physicists are actually discussions of tensor *fields*. A tensor field is a continuous (and usually smooth) assignment of tensors to points on a man... (more) |
— | about 3 years ago |
| Comment | Post #284718 |
Slightly less objectively, your wording and notation compounded by your inconsistent use and unfortunate choices of notation confuse the issue significantly. A key part of this discussion is that all this co-/contra-variant stuff has to do with how we represent vectors (and tensors generally) *with r... (more) |
— | about 3 years ago |
| Comment | Post #284718 |
There are several issues here. First, you call everything a "vector". What the notation refers to are *tensors* and vectors are just the rank-1 case. Anything with other than 1 index is NOT a vector. Your description of the Einstein summation convention is completely wrong. It is not a convention to ... (more) |
— | about 3 years ago |
| Edit | Post #284284 | Initial revision | — | about 3 years ago |
| Answer | — |
A: $\int dx dy dz d p_x dp_y dp_z$ Does it have any physical meaning? It's hard to answer your question specifically without the context, and obviously the physical significance of some expression depends on what the variables and operations in that expression stand for. Before considering this particular integral, I want to talk about integration generally and its not... (more) |
— | about 3 years ago |
| Edit | Post #284283 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Is Pythagorean theorem really valid in higher dimensional space? Consider a vector, $\mathbf v=(a,b,c)$, in $3$-space. We can project this onto the $xy$-plane, say, producing the vector $(a, b, 0)$ which we can identify with the $2$-vector, $(a, b)$. This two vector corresponds to the hypotenuse of a right triangle whose side lengths are $a$ and $b$. Therefore the... (more) |
— | about 3 years ago |
| Edit | Post #283885 |
Post edited: Talk about the actual total derivative |
— | about 3 years ago |
| Comment | Post #283900 |
I haven't downvoted (or upvoted) this meta answer, but I haven't yet completely decided how I feel. On the one hand, I think *in practice* there are indeed a lot of bad questions of the form discussed. On the other hand, I *am* making the case that there are often better answers than "experience, per... (more) |
— | about 3 years ago |
| Comment | Post #283900 |
It is also the case that the need for insight is often oversold. Many results about sums of binomial coefficients and closely related combinatorial identities have clever proofs involving insights and analogies. However, virtually all "textbook" identities of this form fall within the purview of [a d... (more) |
— | about 3 years ago |
| Comment | Post #283900 |
To give an example of the kind of thing I'm talking about, consider the following more positive rendition of the infinitude of primes: For any finite set of primes, there exists a prime not in that set. I can produce the following proof skeleton completely mechanically from the form of the propositio... (more) |
— | about 3 years ago |
| Comment | Post #283900 |
There is actually quite a bit that can be mechanized in proof finding. Up to and including completely deriving a proof mechanically. And I don't just mean that automatic theorem provers exist. A decent amount of a proof is highly constrained. In my experience, it is not uncommon for people to struggl... (more) |
— | about 3 years ago |
| Edit | Post #283947 | Initial revision | — | about 3 years ago |
| Answer | — |
A: What're the orders for equation expressing? This question contains a lot of confusion. First, an equation is something like $f(x) = g(x)$ (i.e. there's an equality sign). Solving an equation means finding (all) values for the free variables such that both sides become equal. In the above example, this would mean finding values for $x$ such tha... (more) |
— | about 3 years ago |
| Edit | Post #283885 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Getting backward of partial differentiation's chain rule Traditional mathematical notation for calculus (both integral and differential) is rather incoherent. I don't think there exists a write-up providing systematic rules that would allow you to correctly and unambiguously parse this kind of notation, i.e. the kind of notation used in a typical undergrad... (more) |
— | about 3 years ago |
| Comment | Post #282900 |
While it's fine to self-answer a question, why are you answering it as if you weren't the person to ask it, e.g. saying "your answer isn't correct" and "you got it wrong"? There is no need to pretend like you are a third party who just happened upon this question. In fact, it's a bit confusing and bi... (more) |
— | over 3 years ago |
| Comment | Post #282780 |
I am talking about definite integrals because integration by parts is usually discussed in that context, and the bounds are often important (and sometimes annoying but critical) in applications. Either way, the definite integral formulation is only *more* informative than an indefinite integral. The ... (more) |
— | over 3 years ago |
| Edit | Post #282780 | Initial revision | — | over 3 years ago |