Activity for Peter Taylorâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #287761 |
Your comment about the unapproved edit possibly points to the need for a feature request. I'm a site moderator but the moderation tools don't seem to show a queue of edits pending approval. If it's still pending, post a link in this comment thread and I'll look at it. (more) |
— | almost 2 years ago |
| Comment | Post #287674 |
The choice of symbols is separate to the choice of base. The symbols which you call Arabic numerals in the question aren't the 10 symbols used in Arabic. Computer programmers tend to use 0123456789abcdef for hexadecimal. But it seems to me that the choice of symbols is a question of communication as ... (more) |
— | almost 2 years ago |
| Comment | Post #287674 |
Is this a question about mathematics? (more) |
— | almost 2 years ago |
| Comment | Post #287588 |
[Quadratic programming](https://en.wikipedia.org/wiki/Quadratic_programming) has a long Wikipedia page which I won't attempt to summarise beyond to say that it's about optimising an "objective function" subject to quadratic constraints. Since all you care about is the existence of a solution, the cho... (more) |
— | almost 2 years ago |
| Comment | Post #287588 |
"Correct assignment of signs": when I said "signed volume" it's because if the given value is negative you need to negate it to get the correct volume. Assigning the signs is essentially guessing which ones need to be negated. (The last sentence gives an idea for making an intelligent guess; in the w... (more) |
— | almost 2 years ago |
| Edit | Post #287588 |
Post edited: |
— | almost 2 years ago |
| Edit | Post #287588 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Dividing a cuboid in four Firstly, I assume that your usage of "quadrilateral" implies planarity, and I will largely assume that the hexahedron is convex. Then the hexahedron can be represented as six vectors, being the (unnormalised) normals for the planes under the convention that the plane is given by $v \cdot N = 1$. This... (more) |
— | almost 2 years ago |
| Comment | Post #287518 |
I think it would be possible to replace `$$` with `$`, similarly with whatever other block markers are configured, and `\begin` with `%` or something else which makes the eqnarray or cases or whatever environment obviously not render so that the poster realises there's something to fix. (more) |
— | almost 2 years ago |
| Edit | Post #287519 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Posts with math are very, very tall in the feed Apart from my other answer about titles, which is posted separately so that people can vote on it separately, I don't see a problem which needs fixing. (more) |
— | almost 2 years ago |
| Edit | Post #287518 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Posts with math are very, very tall in the feed I think that forcing inline display of a given MathJax segment in some contexts but not others is not a great solution because there are a few things which get treated differently in the two contexts, to the extent that I change some markup if I decide to change a block to an inline block. However... (more) |
— | almost 2 years ago |
| Edit | Post #287410 |
Post edited: When I fixed the consistency in the body earlier I overlooked the title |
— | almost 2 years ago |
| Edit | Post #287502 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Is posting multiple answers encouraged, and under what circumstances? I'm not aware of any specific guidance, so for now it probably comes down to "Use your common sense". Certainly if the answers are long and independent then it makes sense to post them separately; if they're all one-liners then it probably makes more sense to post them in one answer. (more) |
— | almost 2 years ago |
| Comment | Post #287481 |
View Source shows that the raw HTML being sent to the browser is `A parabola is given by $y^2=2px$ with $p>0$. The point $D$ is on the parabola in the first quadrant at a distance of $8$ from the $x$-a...` so the problem is a double-escaping. (more) |
— | almost 2 years ago |
| Edit | Post #287410 |
Post edited: Make the two expressions consistent with each other, use more standard limits |
— | almost 2 years ago |
| Comment | Post #287492 |
Rather than mapping the whole original example, maybe you can split it: at least one of the ranges $[0, 1]$ and $[1, \infty)$ must preserve the property, and if it's the latter then it's at least worth considering that it still preserves the property under the mapping $x \to \frac1x$. (more) |
— | almost 2 years ago |
| Edit | Post #287439 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Find all integer solutions for $a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = n^2$ The initial observation is explained by the identity $$(x^2 + 3x + 1)^2 = x(x+1)(x+2)(x+3) + 1$$ The generalisation is messed up in the question as it currently stands, but the only smallish similar identity which I can find is the uninteresting $$(x+1)^2 = x(x+2) + 1$$ (more) |
— | almost 2 years ago |
| Comment | Post #287178 |
@#55348, if you take any 2D shape with a well-defined area and sweep it along its normal to create a generalised prism of height h, the volume will be the area times h. If you linearly scale the shape by the same scale in both directions, the area of the shape (and hence the volume of the generalised... (more) |
— | about 2 years ago |
| Edit | Post #287146 | Initial revision | — | about 2 years ago |
| Answer | — |
A: What to do with the algebra-precalculus tag? After a week, I've taken action in line with the feedback received by eliminating the tag, retagging some of the questions which had it, and deleting some which had no redeeming features at all (negative score, no upvotes, no answers except for a self-answer by a deleted user). (more) |
— | about 2 years ago |
| Edit | Post #282966 |
Post edited: Eliminating unhelpful tag algebra-precalculus |
— | about 2 years ago |
| Edit | Post #280118 |
Post edited: Eliminating unhelpful tag algebra-precalculus |
— | about 2 years ago |
| Edit | Post #286140 |
Post edited: Eliminating unhelpful tag algebra-precalculus |
— | about 2 years ago |
| Edit | Post #287063 |
Post edited: Eliminating unhelpful tag algebra-precalculus |
— | about 2 years ago |
| Edit | Post #287078 | Initial revision | — | about 2 years ago |
| Question | — |
What to do with the algebra-precalculus tag? This question is prompted by a recent question with the tag algebra-precalculus, although I see that it's by no means the first. This tag apparently means something to some people. I believe that specifically it means something to people who went through the US education system (whether in the US ... (more) |
— | about 2 years ago |
| Comment | Post #286991 |
I've seen this problem before, I but can't remember the details. One thing which does stand out is that the extrema settle. If you consider e.g. ordering $1,2,5,3,4,6$, the $1,2,\ldots$ and $\ldots,6$ are now fixed. (more) |
— | about 2 years ago |
| Comment | Post #286854 |
Ok, I get it now. For some reason I was visualising the spiral as going inwards, not outwards. (more) |
— | over 2 years ago |
| Comment | Post #286854 |
This isn't bijective: it doesn't encode information about the width and height of the spiral, so it can't be decoded. (more) |
— | over 2 years ago |
| Comment | Post #286709 |
I wonder whether the textbook was badly translated into English. A more standard name for $m_1, m_2, \ldots, m_k$ would be *intermediate terms*, and you'll probably find that more useful for communicating with people who didn't use the same textbook. (more) |
— | over 2 years ago |
| Comment | Post #286709 |
Are you sure that "mean" is the correct word? The arithmetic mean of a collection of numbers is the most common form of average. I would guess from the example that the term you want is "number of intermediate values", but I'm not certain that it's the only possibility. (more) |
— | over 2 years ago |
| Edit | Post #286656 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Do the Faber partition polynomials have integer coefficients? Let $B(x) = b1 x + b2 x^2 + \cdots$. Then $$\begin{eqnarray} Fn(b1, \ldots, bn) &=& - [x^n] n \log(1 + B(x)) \\\\ &=& [x^n] n \sum{i \ge 1} \frac{(-B(x))^i}{i} \\\\ &=& \sum{\lambda \\, \vdash \\, n} \frac{n}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f1, \ldots, fn} \prodj ... (more) |
— | over 2 years ago |
| Edit | Post #286655 | Initial revision | — | over 2 years ago |
| Question | — |
Do the Faber partition polynomials have integer coefficients? The Online Encyclopedia of Integer Sequences includes A263916: Coefficients of the Faber partition polynomials. Perhaps the clearest definition given is > -log(1 + b(1) x + b(2) x^2 + ...) = Sum{n>=1} F(n,b(1),...,b(n)) x^n/n which in better notation is $$\sum{n \ge 1} Fn(b1, \ldots, bn) \f... (more) |
— | over 2 years ago |
| Comment | Post #286569 |
I tried two cost functions for "even spacing". If the *gap* between two consecutive values is the difference of their logarithms, the first cost function was the difference between the largest gap and the smallest gap; and the second cost function was the variance of all the gaps. I can't tell you of... (more) |
— | over 2 years ago |
| Comment | Post #286572 |
I don't know and right now can't see the answer, but the obvious point (which has probably already occurred to you but which I make for completeness) is that you don't seem to have used the precondition "$\alpha \odot \beta$ is the identity automorphism". (more) |
— | over 2 years ago |
| Edit | Post #286569 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286569 |
Post edited: Investigate with some simple cost functions |
— | over 2 years ago |
| Edit | Post #286569 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Optimising a 3 value problem (well 2 really) > This is the list of combinations I have found, the result size sequence changes depending on the three values so only the first and last four are always in the same place. > > ``A||B||C, A||B, A||C, A||(B+C), A, B||C, (A+C)||B, (A+B)||C, B, A+B||C, A||C+B, A+B, C, A||B+C, A+C, B+C, A+B+C`` I t... (more) |
— | over 2 years ago |
| Comment | Post #286453 |
I can't see where the conclusion $D_x g^{-\epsilon} < 0$ comes from. $\epsilon > 0$, $g^{1+e}(x) > 0$, $g'(x) > 0$, so it seems to me that $D_x g^{-\epsilon} > 0$. (more) |
— | over 2 years ago |
| Edit | Post #286453 |
Post edited: Improve legibility |
— | over 2 years ago |
| Comment | Post #286472 |
Two things. 1. For consistency with the other tags on the site, please use lower case unless there's a good reason not to (e.g. people's names, standard upper-case symbols). 2. What is "numerical" intended to capture? Numerical analysis? (more) |
— | over 2 years ago |
| Edit | Post #286399 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Matrices with rotational symmetry Notation: $A^{\leftarrow}$ denotes $A$ with the columns reversed; $A^{\uparrow}$ denotes $A$ with the rows reversed; $A^{\leftarrow \uparrow} = A^{\uparrow \leftarrow}$ is denoted $A^{\circ}$ and is the rotation of $A$ by $180^{\circ}$. Consider first a $(2n+1)\times(2n+1)$ block matrix $\begin{pm... (more) |
— | over 2 years ago |