Activity for Derek Elkinsâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #287879 |
Post edited: |
— | about 1 year ago |
| Edit | Post #287879 | Initial revision | — | about 1 year ago |
| Answer | — |
A: Generalization of categorical product The $F(\pii) = id$ doesn't really fit the form of a universal property. If we drop that constraint, we can present your universal property in terms of representability $$\mathcal C(Y, X1 \timesF X2) \cong \{(f1,f2)\in\mathcal C(Y, X1)\times\mathcal C(Y, X2)\mid F(f1) = F(f2) \}$$ natural in $Y$. No... (more) |
— | about 1 year ago |
| Edit | Post #287849 |
Post edited: Improve kerning of quotient set notation. |
— | about 1 year ago |
| Comment | Post #287849 |
Both here and in your answer you are not explicitly using the fact that $f$ is constant on equivalences classes nor the definition of the quotient topology. You implicitly use the former when saying $g([x]):=f(x)$ is well-defined in the vague sketch of the proof at the beginning, but not in the separ... (more) |
— | about 1 year ago |
| Suggested Edit | Post #287849 |
Suggested edit: Improve kerning of quotient set notation. (more) |
helpful | about 1 year ago |
| Edit | Post #287721 | Initial revision | — | over 1 year ago |
| Answer | — |
A: Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem By definition $\mathbb R^\mathbb R \subset \mathbf 2^{\mathbb R\times\mathbb R}$. So all we need is a surjection $\mathbb R \twoheadrightarrow \mathbb R \times \mathbb R$ of which there are plenty such as space filling curves. If you have a bijection between $\mathbf 2^\mathbb N$ and $\mathbb R$, the... (more) |
— | over 1 year ago |
| Edit | Post #287644 | Initial revision | — | over 1 year ago |
| Answer | — |
A: What is the Name of Function for Probability of a Certain Sum on Random Die Rolls? I don't know some special name for exactly that sum, but it is closely related to what are known as polynomial coefficients or extended binomial coefficients. These are often written as ${n \choose j}{k+1}$ and are defined indirectly via: $$\left(\sum{i=0}^k x^i\right)^n = \sum{j=0}^{nk} {n \choose j... (more) |
— | over 1 year ago |
| Comment | Post #287625 |
The equation you give in the title doesn't seem to appear in the image (which you should reproduce the relevant part as MathJax and also crop it much more aggressively). The equation you reference is $\int_C f(x,y)ds = \int_a^b f(x, 0)dx$ There is no $f(x)$. The domain of integration is critical here... (more) |
— | over 1 year ago |
| Edit | Post #287003 | Initial revision | — | over 1 year ago |
| Answer | — |
A: Proving that $p\mid (p+1776)$ if $p$ is a prime and $p(p+1776)$ is a perfect square You basically have it. I'll give an overly detailed rendition of an informal proof below. As you state, the assumption is that $p(p+1776) = k^2$ for some natural number $k$ with $p$ a prime. We want to show that $p\mid p + 1776$, or, equivalently, $p + 1776 = pm$ for some natural $m$. Proof:... (more) |
— | over 1 year ago |
| Edit | Post #286957 |
Post edited: cases environment accomplishes the same output in a simpler way |
— | over 1 year ago |
| Suggested Edit | Post #286957 |
Suggested edit: cases environment accomplishes the same output in a simpler way (more) |
helpful | over 1 year ago |
| Comment | Post #286908 |
@#8056 That's also a common definition according to Wikipedia, though [Wikipedia seems to agree with my experience](https://en.wikipedia.org/wiki/Exponentiation#Real_exponents) that the logarithm-based definition is more common. The logarithm-based definition seems more convenient and easier to work ... (more) |
— | over 1 year ago |
| Comment | Post #286908 |
For $e^x$, given *some other* definition of $e^x$, you can certainly prove that the power series definition I gave is the Taylor series. However, if you take the power series I gave as the definition, then there's no need to talk about the theory of Taylor series. You just have a power series; the fa... (more) |
— | over 1 year ago |
| Comment | Post #286908 |
$\lim_{a\to 0}a$ is just $0$. The limit of an (real) expression is either undefined or it's a real number. It's not some infinitesimal thing. So $\forall x.P(x)\implies P(x\pm\lim_{\epsilon\to 0}\epsilon)$ is just $\forall x.P(x)\implies P(x)$. You could try to do something like $\forall x.\exists\ep... (more) |
— | over 1 year ago |
| Edit | Post #286907 |
Post edited: Format implications so it not an unreadable blob of text. Correct minor typo/thinko. |
— | over 1 year ago |
| Edit | Post #286908 | Initial revision | — | over 1 year ago |
| Answer | — |
A: Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction You can't do induction on the real numbers because the real numbers aren't inductively defined. That said, it's not clear what you intend "induction on reals" to mean. Before continuing, I'd like to make an aside on proof writing. First, as a matter of presentation, it's useful to make it clear wh... (more) |
— | over 1 year ago |
| Suggested Edit | Post #286907 |
Suggested edit: Format implications so it not an unreadable blob of text. Correct minor typo/thinko. (more) |
helpful | over 1 year ago |
| 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 1 year ago |
| Edit | Post #286833 |
Post edited: |
— | almost 2 years ago |
| Edit | Post #286833 | Initial revision | — | almost 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) |
— | almost 2 years ago |
| Edit | Post #286174 |
Post edited: fix typos |
— | almost 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) |
— | almost 2 years ago |
| Edit | Post #286482 | Initial revision | — | almost 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) |
— | almost 2 years ago |
| Edit | Post #286174 |
Post edited: Add some additional historical context |
— | almost 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) |
— | about 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) |
— | about 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) |
— | about 2 years ago |
| Edit | Post #286174 | Initial revision | — | about 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) |
— | about 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) |
— | about 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) |
— | about 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) |
— | about 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) |
— | about 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) |
— | about 2 years ago |
| Edit | Post #285978 | Initial revision | — | about 2 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) |
— | about 2 years ago |
| Edit | Post #285442 | Initial revision | — | over 2 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) |
— | over 2 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) |
— | over 2 years ago |
| Edit | Post #285424 |
Post edited: Fix braces. |
— | over 2 years ago |
| Suggested Edit | Post #285424 |
Suggested edit: Fix braces. (more) |
helpful | over 2 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) |
— | over 2 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) |
— | over 2 years ago |
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