Activity for Derek Elkins
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #280866 |
If I wrote $g\nabla f$, this would mean $g$ times $\nabla$ applied to $f$. For $g$ scalar-valued, if I also wanted to apply $\nabla$ to $g$, I could write $\nabla(gf)$. If $g$ isn't scalar-valued, then even at the algebraic level (i.e. ignoring the differentiation aspect) $g$ doesn't necessarily ... (more) |
— | 5 months ago |
| Comment | Post #296026 |
@#118034 You may find [WHAT DOES IT TAKE TO PROVE FERMAT’S LAST THEOREM?](https://doi.org/10.2178/bsl/1286284558) [PDF](https://www.cs.umd.edu/users/gasarch/BLOGPAPERS/fltlargecard.pdf) even if you were only using FLT as an example. This is not a technical paper but more a discussion around the d... (more) |
— | 5 months ago |
| Edit | Post #295536 | Initial revision | — | 7 months ago |
| Answer | — |
A: How can one prove Tychonoff's theorem using ultrafilters? Presumably, the approach is the one mentioned on Wikipedia, specifically the third proof by Cartan and Bourbaki. It is summarized there, under the assumption, which requires the Ultrafilter Lemma, that a topological space is compact if and only if all ultrafilters on that space converge. Given... (more) |
— | 7 months ago |
| Edit | Post #295147 | Initial revision | — | 9 months ago |
| Answer | — |
A: What explains the Inventor’s Paradox? Why can more general problems, paradoxically, be easier to solve or prove? There are many reasons why generalizing a problem can make finding a solution simpler. Of course, this is just a heuristic. Generalizing a problem can easily make it much harder or even impossible to find a solution. Here are several reasons why this is the case in decreasing order of objectiv... (more) |
— | 9 months ago |
| Comment | Post #295130 |
The term "leading question" suggests that the one asking has an intended answer that the question is leading toward. If you are trying to *define* a notion you're calling a "leading question" then 1) this seems like a bad name, 2) why link to Wikipedia which isn't using your definition, and 3) wh... (more) |
— | 9 months ago |
| Comment | Post #295130 |
As your link to Wikipedia indicates, "leading question" is not a mathematical term. What are you trying to accomplish with this notion of "leading question"? Why can't you just put forward the notion of measure or whatever that you want and prove that it satisfies the desired properties. Also pro... (more) |
— | 9 months ago |
| Edit | Post #295006 |
Post edited: |
— | 10 months ago |
| Edit | Post #295006 |
Post edited: |
— | 10 months ago |
| Edit | Post #295006 | Initial revision | — | 10 months ago |
| Answer | — |
A: How to justify: for every integer $r$ in $[1, k-1]$, there is an integer $j$ in $[0, k-1]$, such that $r + j = k$ This is a common problem when you transition from presentations of mathematics for a general audience to presentations aimed at potential mathematicians. You realize that many of the concepts you've been taking for granted for years were never actually clearly defined. Concepts like natural numbe... (more) |
— | 10 months ago |
| Comment | Post #294117 |
What this corresponds to in index notation is summing over repeated indices where one is lowered and the other raised, i.e. exactly what Einstein convention covers. Since $U$, $V$, and $W$ can all themselves be tensors, this corresponds to summing over multiple pairs of raised and lowered indices... (more) |
— | over 1 year ago |
| Comment | Post #294117 |
What you *can* talk about is tensor contraction (and various other much less common operations that seem even further from what you want). This corresponds to the fact that the identity linear transformation in $\mathrm{Lin}(V,V)$ gives rise to a linear transformation $\mathrm{Lin}(V\otimes V^\*,... (more) |
— | over 1 year ago |
| Comment | Post #294117 |
The reason I said the fact that composition of linear transformations don't form a group was a "minor nit" was because we can talk about a unit to composition without inverses. In technical terms, linear transformations from and to the same vector space form a monoid. More generally, vector space... (more) |
— | over 1 year ago |
| Comment | Post #294117 |
The reason we can view order-2 tensors as matrices is because $\mathrm{Lin}(U,V)\cong\mathrm{Lin}(U\otimes V^\*,\mathbb R)$, and we can view an element of $\mathrm{Lin}(U\otimes V^\*,\mathbb R)\cong\mathrm{MultiLin}(U,V^\*;\mathbb R)$ as an order-2 tensor, i.e. a multilinear form. The dual vector... (more) |
— | over 1 year ago |
| Comment | Post #293347 |
I assume you're assuming $f$ is analytic, i.e. can be defined by a convergent Taylor series, otherwise it's easy to come up with counterexamples. Furthermore, I assume you're assuming $f$ is invertible, e.g. the $\sin$ function is analytic but clearly many values map to $0$. Alternatively, you'd ... (more) |
— | over 1 year ago |
| Comment | Post #293102 |
We're unioning sets of axioms, not combining models. Having two models that give different assignments to propositional variables (which is what I meant by "conflicting") doesn't mean that there's a contradiction. Even for the same set of axioms we can have multiple models, e.g. there are at leas... (more) |
— | almost 2 years ago |
| Comment | Post #293102 |
To answer the question in your first footnote (which seems to be missing some words, but I believe the intent is clear enough): The statement says what it says, which is that each finite subset has a model. There is no assumption that there is one model that will work for every finite subset, tho... (more) |
— | almost 2 years ago |
| Edit | Post #293102 |
Post edited: TeXify. Remove set-theory tag. Change footnotes into real footnotes, which may not be an improvement. Break the text into paragraphs, though it can probably be done better or more in line with the original source. |
— | almost 2 years ago |
| Suggested Edit | Post #293102 |
Suggested edit: TeXify. Remove set-theory tag. Change footnotes into real footnotes, which may not be an improvement. Break the text into paragraphs, though it can probably be done better or more in line with the original source. (more) |
helpful | almost 2 years ago |
| Comment | Post #293102 |
This looks like a compactness theorem, not a completeness theorem. (more) |
— | almost 2 years ago |
| Edit | Post #293013 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: How to validate if the horizontal and vertical tangent lines exist for implicit functions? Answering your questions a bit out of order, I'll start with the "non-rigorousness" of talking about $dx$ by itself. While this seems to be commonly poorly explained, the derivative (of a function from and to reals) is an operation that takes functions to functions. Let's consider the typical hig... (more) |
— | almost 2 years ago |
| Edit | Post #292410 |
Post edited: fix typo I introduced, cap -> cup, tweak title, remove symm tag |
— | about 2 years ago |
| Suggested Edit | Post #292410 |
Suggested edit: fix typo I introduced, cap -> cup, tweak title, remove symm tag (more) |
helpful | about 2 years ago |
| Edit | Post #292410 |
Post edited: Correct typo and formatting errors, also ^{\prime} -> ' |
— | about 2 years ago |
| Edit | Post #292418 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Why $\gamma\cdot\operatorname{grad}u<0$ in the Theorem? (Nirenberg academic paper) As the proof of Theorem 2 suggests, this follows immediately from Theorem 2.1. While it's a bit ambiguously worded, to apply Theorem 2.1, we need $b1(x)=0$ in $\Delta u + b1(x)u{x1} + f(u) = 0$, $u > 0$ in $\Omega$, $u = 0$ on a part of $\partial\Omega$, and some basic continuity conditions. The ... (more) |
— | about 2 years ago |
| Suggested Edit | Post #292410 |
Suggested edit: Correct typo and formatting errors, also ^{\prime} -> ' (more) |
helpful | about 2 years ago |
| Edit | Post #292320 |
Post edited: |
— | about 2 years ago |
| Edit | Post #292320 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Reflection in the plane with polar coordinates Perhaps the simplest way to get the formula is to think geometrically. Let's say we wanted to reflect a point $x=(r\cos\theta,r\sin\theta)$ across the $x$-axis. In that case, we can simply negate $\theta$ giving $(r\cos(-\theta),r\sin(-\theta))=(r\cos\theta,-r\sin\theta)$ as expected. If we... (more) |
— | about 2 years ago |
| Edit | Post #292225 |
Post edited: Formatting and minor typo fixes and grammar tweaks. Backslashes sometimes need to be escaped leading to things like \\\\ for \\. |
— | about 2 years ago |
| Edit | Post #292231 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Complex functions and inner product $\langle \frac{\partial f}{\partial z} , g\rangle $ There are multiple issues with how you compute the exponent of $r$. The first issue is you compute $r^{j-1}r^k = r^{j+k}$ rather than $r^{j+k-1}$. You compound this error when you substitute $j = k+1$ into $r^{j+k}$ (which should be $r^{j+k-1}$) in the formula for $\left\langle\frac{\partia... (more) |
— | about 2 years ago |
| Suggested Edit | Post #292225 |
Suggested edit: Formatting and minor typo fixes and grammar tweaks. Backslashes sometimes need to be escaped leading to things like \\\\ for \\. (more) |
helpful | about 2 years ago |
| Edit | Post #292202 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Find the value of $\sum_{k=1}^\infty\frac{k^2}{k!}$ This answer is, in some ways, "just" a rephrasing of the power series answer by Snoopy, but the broader perspective and name-dropping the relevant tools may be useful. The relevant tool being generating functions. While not the best book on the topic, Herbert Wilf's generatingfunctionology is ... (more) |
— | about 2 years ago |
| Comment | Post #292189 |
$\lg$ (versus $\log$) is commonly used for the base 2 logarithm, though it is certainly isn't unambiguous. I've never seen $\operatorname{ld}$ or $\operatorname{lb}$ used for this, so it seems like a terrible choice for clarity. $\log_2$ is, of course, unambiguous. (more) |
— | about 2 years ago |
| Comment | Post #291919 |
Answering for Peter Taylor, it's a rotation matrix. You could derive it, as you could any matrix, by considering a rotation operator and computing where it sends basis vectors. In this case, you could compute that the vector $(1,0)$ gets sent to $(\cos\alpha,\sin\alpha)$ by a (counter-clockwise) ... (more) |
— | about 2 years ago |
| Edit | Post #291823 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Interpreting $\text{Prop}$ in Set Not asserting an axiom only broadens the class of possible models. Unless you're adding an axiom that states that the type theory is definitively not proof irrelevant, any proof irrelevant model is still a model. Whether or not proof irrelevance is assumed would only matter if it excluded some mo... (more) |
— | about 2 years ago |
| Edit | Post #291617 |
Post edited: Make more legible and grammatical. |
— | over 2 years ago |
| Comment | Post #291588 |
Maybe I don't understand what you intend with the notation, but ${+}^{-1}(U)$ is a subset of $V \times V$ and so definitely isn't in $\mathcal T_k$. (more) |
— | over 2 years ago |
| Comment | Post #291617 |
$\alpha$ and $\beta$ need to be homomorphisms of the entropic structure. Presumably, this means $\alpha(x \odot y) = \alpha(x) \odot \alpha(y)$. $x^2$ doesn't seem like a homomorphism for the entropic structure you defined. That is, $(x + 2y)^2 = x^2 + 4xy + 4y^2 \neq x^2 + 2y^2$ even mod $3$. (more) |
— | over 2 years ago |
| Suggested Edit | Post #291617 |
Suggested edit: Make more legible and grammatical. (more) |
helpful | over 2 years ago |
| Edit | Post #291652 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Why does the method of separating variables work? I don't think there's a satisfying answer to this question currently. The very first problem – which is probably surprising – is that there isn't a widely accepted, general definition of "separation of variables". An obvious approach to studying separation of variables would be to ... (more) |
— | over 2 years ago |
| Comment | Post #291500 |
I simply substituted in the value for $\beta$. (more) |
— | over 2 years ago |