Activity for Peter Taylor
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #284561 |
Post edited: Remove tags which don't seem likely to fit many questions and create one which does |
— | about 3 years ago |
| Comment | Post #284561 |
If you're asking about notations that physicists use, you might have a better chance of getting a good answer on the physics site. (more) |
— | about 3 years ago |
| Edit | Post #284712 |
Post edited: |
— | about 3 years ago |
| Edit | Post #284712 |
Post edited: |
— | about 3 years ago |
| Edit | Post #284712 | Initial revision | — | about 3 years ago |
| Answer | — |
A: ratio of partial sums of the same geometric sequence I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 \approx \frac{819}{51} \approx 16.05$$ and th... (more) |
— | about 3 years ago |
| Comment | Post #284651 |
You can easily calculate the expected steps per roll as 1.68. To reach a million steps you're going to need at least 20000 rolls, and an estimated 595238 rolls. Is there anything which you can't calculate to sufficient precision for your purposes just using the central limit theorem? (more) |
— | about 3 years ago |
| Comment | Post #284550 |
`$\tilde{e}$` gives $\tilde{e}$ (more) |
— | about 3 years ago |
| Comment | Post #284052 |
You seem to be amalgamating two orthogonal issues: citing the source of the material which the question is about, and establishing some kind of whitelist of acceptable sources. IMO the first is a reasonable issue to raise and the second is an unreasonable suggestion, but the discussion would be clear... (more) |
— | about 3 years ago |
| Comment | Post #283952 |
Yes, that's what I said in my comment yesterday. (more) |
— | about 3 years ago |
| Edit | Post #283111 | Question closed | — | about 3 years ago |
| Comment | Post #283935 |
@#53922 I think (although without more context I can't be certain) that you're misunderstanding the flow of the text. It looks to me as though (2.6) states a condition (which you've misquoted in your comment: that subscript $\alpha = 0$ is there for a reason) and then the following text, from "By the... (more) |
— | about 3 years ago |
| Edit | Post #283935 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Consider the second of these integrals (What's the meaning of second right here?) > What did they mean by "second"? They've mentally expanded $$\frac{dJ}{d\alpha}=\int{x1}^{x2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\right)\mathrm dx$$ (which I've corrected to be what it say... (more) |
— | about 3 years ago |
| Comment | Post #282564 |
@#54204, I'm not sure what suggested edits you're talking about. (more) |
— | about 3 years ago |
| Comment | Post #283886 |
I don't think it's a duplicate as such, but you asked a [previous question](https://math.codidact.com/posts/282771) on a later sentence from this exact passage from your textbook (and Wolgwang kindly replaced your image with text and MathJax there, which you could reuse here). (more) |
— | about 3 years ago |
| Comment | Post #283899 |
@DNB, your teacher seems to have a vocabulary full of obscure English words which reference supernatural beings or supernatural knowledge of the future (*soothsay*, *augure*, *fey*, *sibylline*, *vaticinate*), but that vocabulary isn't suitable for use (a) in mathematics; (b) with native English spea... (more) |
— | about 3 years ago |
| Edit | Post #283713 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Find limits of integration in polar coordinates You're going to need to differentiate the curves where they pass through the origin, and also find the angle of the intercept. $$\frac{\textrm{d}}{\textrm{d}x} \textrm{blue curve}\Big\vert{x=0} = 0$$ giving a limit $\theta0=0$. $$\frac{\textrm{d}}{\textrm{d}x} \textrm{red curve}\Big\vert{x=0} =... (more) |
— | about 3 years ago |
| Comment | Post #283400 |
Hint: you can find the answer near the start of the [Wikipedia page on matrices](https://en.wikipedia.org/wiki/Matrix_\(mathematics\)). (more) |
— | over 3 years ago |
| Edit | Post #283400 |
Post edited: Escaping workaround |
— | over 3 years ago |
| Edit | Post #283518 | Initial revision | — | over 3 years ago |
| Question | — |
Do these major triangle lines have names? Kimberling's Encyclopedia of Triangle Centers lists, among other things, lines on which each centre is found, but usually listing only two points on the line. As a little project I've assembled these triples to find lines which have many centres, which would seem to be a rough measure of how importan... (more) |
— | over 3 years ago |
| Comment | Post #283399 |
@#53628, it's not an unreasonable assumption that someone with questions about material at this level is in full-time education and has access to a teacher. (more) |
— | over 3 years ago |
| Edit | Post #283433 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Finding the smallest Mersenne-number multiple of an odd integer Add one to both sides and consider residues modulo $a$ to get $$2^n \equiv 1 \pmod a$$ So you want to find the multiplicative order of $2$ modulo $a$. As you note, $2^n \ge a + 1$ so $n \ge \lg(a+1)$; by Lagrange's theorem, $n \le \varphi(a)$, where $\varphi$ is Euler's totient function. More gene... (more) |
— | over 3 years ago |
| Comment | Post #283339 |
You do realise that this is the Mathematics site, not the Physics one? (more) |
— | over 3 years ago |
| Comment | Post #283299 |
Yes, and then we can see how well the "revert to previous version" feature is implemented. (more) |
— | over 3 years ago |
| Edit | Post #283301 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Is replacing the entire question with a different one appropriate? No, this is not appropriate. If someone no longer wishes to keep their unanswered question on the site, they can delete it. If someone wishes to ask a new question, they can do so as a new question. The only reasons I can see for not behaving straightforwardly are (i) some idea that it's more ecol... (more) |
— | over 3 years ago |
| Edit | Post #283299 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why was my question closed: If Alice must've have classes on at least 2 days, why do you need the intersection of 3 's? 1. It wasn't actually unilateral. 2. It's not so much that the question was unconstructive, as the wholesale replacement of the content which removed all context to the existing comment. The question was flagged as "seems like some weird attempt to game the system", which indeed it does. So I pos... (more) |
— | over 3 years ago |
| Comment | Post #283254 |
Does this really need two questions? There are marginal differerences to the other one, but really they're both asking for explanations of the same diagram, and if there's any answer other than "If that explanation doesn't help you, find a better one" then it's probably better in one place rather tha... (more) |
— | over 3 years ago |
| Edit | Post #282645 | Question closed | — | over 3 years ago |
| Edit | Post #283185 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Prove $(\cos^3\theta+\sin^3\theta)^2= \cos^6\theta(1+\tan^3\theta)^2$ $$(a+b)^2 = a^2 \left(1 + \frac ba\right)^2$$ (more) |
— | over 3 years ago |
| Comment | Post #282600 |
What makes you think that "*these authors and publishers are desperate for income*" as opposed to unsatisfied with the alternatives? (more) |
— | over 3 years ago |
| Edit | Post #283122 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Intuitively, why would organisms — that after one minute, will either die, split into two, or stay the same, with equal probability — all die ultimately? This can be recast as a random walk on a line. Let $nt$ be the number of amoebae after $t$ events, and process the events in any order which makes sense. (It may help to think of this as serialising a parallel process on a single-core CPU). For example, you could choose to number the the amoebae by t... (more) |
— | over 3 years ago |
| Edit | Post #283121 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why would skyrocketing the numbers of doors help laypeople intuit the Monty Hall Problem? The only thing special about the door you chose is that you chose it, and you did so without any information, so objectively it isn't special at all. The door which the host leaves closed is special because it was chosen from the remaining doors by someone with information, so objectively it is actua... (more) |
— | over 3 years ago |
| Edit | Post #279044 |
Post edited: A proposed edit tried to improve the typography with nbsp; this is a better fix |
— | over 3 years ago |
| Comment | Post #283086 |
Firstly, I don't see anything in the T&C on the physicsforums site which says that content is CC0-compatible, which is why I've deleted your self-answer. But secondly, even if it were, it's almost always better to write an answer in your own words once you've understood the solution to a more than su... (more) |
— | over 3 years ago |
| Comment | Post #282645 |
This question at present is completely different to the question originally posted and to which my previous comment applies. Recycling a question ID like that is a source of confusion. What's going on? (more) |
— | over 3 years ago |
| Comment | Post #283086 |
That sounds very broken. DuckDuckGo in an incognito browser window gives me a useful result from Wolfram MathWorld as the very first result. (more) |
— | over 3 years ago |
| Comment | Post #283086 |
What does your favourite search engine say? (more) |
— | over 3 years ago |
| Comment | Post #282658 |
@#8046 , go ahead. (more) |
— | over 3 years ago |
| Comment | Post #282658 |
I won't pretend to be enthusiastic about the idea, but I recognise that in small communities it's sometimes necessary to serve a term in office as a public duty. I have no prior experience as a moderator *per se*, but in another place I did have maximum rep-based privileges unlocked on one site and m... (more) |
— | over 3 years ago |
| Comment | Post #282642 |
If the quoted exercise is the true problem and this is an XY question, it's far simpler to consider the basic combinatorial meaning of $\binom{n}{k}$. (more) |
— | over 3 years ago |
| Comment | Post #282645 |
Probably not. But if you interpret $\sum_{k=0}^n k \binom{2n}{k}$ in terms of choosing a team of up to $n$ people with one designated captain from $2n$ people then you can transform it into a sum which you're already familiar with. (more) |
— | over 3 years ago |