Activity for Peter Taylor
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #292896 | Initial revision | — | about 2 months ago |
| Answer | — |
A: Should there be more than one sort of math community? It's not always easy to draw the line. I read this question just now because it was bumped, and immediately thought of this question. It was, at least in part, a question about philosophy of mathematics, but I don't think the asker realised that. Their goal was to understand the context of a paper. ... (more) |
— | about 2 months ago |
| Comment | Post #292671 |
Two independent points: 1. "Convex hull" suggests an underlying set which you're bounding. The description of your data would fit better with the term "convex polytope", which may help you search the literature. 2. "the edge... normal to the vector $\vec{G}_j$" suggests that you're working in 2D. If ... (more) |
— | 3 months ago |
| Edit | Post #292629 | Question closed | — | 3 months ago |
| Comment | Post #292629 |
It's not reasonable to expect people on this website to define terms which appear to have been invented by someone on another website when we can't even read the original post without creating an account. In addition, the real question here appears to be non-mathematical: how can I tell my boss that ... (more) |
— | 3 months ago |
| Comment | Post #292332 |
I think there may be an interesting question in here somewhere, but it needs some work to bring it out. My first impression when reading it was that it is rather confusing. Who are Barsky and Benzaghou, and why is it relevant that they in particular haven't proved something that no-one has proved? Wh... (more) |
— | 4 months ago |
| Edit | Post #292317 |
Post edited: A few small translations from Spanish |
— | 4 months ago |
| Comment | Post #292268 |
Doesn't the hint also require the unstated assumption that $0 \in \Omega$? (more) |
— | 4 months ago |
| Comment | Post #292178 |
I checked the top-left of the table, reading antidiagonals in both directions, and found nothing in OEIS. This is mildly surprising, but it rules out some obvious guesses. (more) |
— | 5 months ago |
| Comment | Post #292112 |
On the positive side, it's sufficient to show that $f(p) = p$ for all primes $p$. If so then for each $1 \le a < p$ there is a prime $q = a + mp$ by Dirichlet's theorem on arithmetic progressions: then $f(q) \equiv a \pmod p$ and since each equivalence class maps to a single equivalence class this me... (more) |
— | 5 months ago |
| Comment | Post #292112 |
It gets messy trying to push it. $f(5) \equiv 5 \pmod {12}$ follows from previous observations; $f(5) \in \\{0, 4\\} \pmod 5$ follows because the other equivalence classes are already accounted for. Then $f(5) - 1$ must be a power of two (it must be $5$-smooth and coprime to $3$ and $5$); $f(5) - 3$ ... (more) |
— | 5 months ago |
| Comment | Post #292112 |
Then the original property is equivalent to the pair of properties: $\forall p \textrm{ prime}: f(a) \equiv f(b) \pmod p \iff a \equiv b \pmod p$ and $\forall p \textrm{ prime}: f(p) \equiv 1 \pmod {p-1}$. By Dirichlet's theorem on arithmetic progressions we have that for every prime $p$ there are an... (more) |
— | 5 months ago |
| Comment | Post #292112 |
Actually we can argue that for each prime $p$ there is a $c_p$ such that $f(p) = c_p(p-1) + 1$. If $f(n) \bmod p$ takes fewer than $p$ distinct values then by the pigeonhole principle there must be $a \not\equiv b$ for which $f(a) \equiv f(b)$ and we can derive a contradiction from $f(a)^{f(p)} \equi... (more) |
— | 5 months ago |
| Comment | Post #292112 |
$f(a)^{f(p)}\equiv f(a)\pmod p$ is satisfied if (but not only if) $f(p) \equiv 1 \pmod{p-1}$, i.e. $f(p) = c_p (p-1) + 1$. But there's no reason *a priori* why $c_p$ should be the same for all primes, or why $f(n)$ should be $c(n-1) + 1$ for non-prime $n$. I think you've only considered a negligible ... (more) |
— | 5 months ago |
| Comment | Post #292113 |
What I see in the browser console is a 404 for amsmath.js. Does the post actually rely on amsmath? Maybe removing that `require` is the solution. (more) |
— | 5 months ago |
| Edit | Post #292062 |
Post edited: Fix typos |
— | 5 months ago |
| Comment | Post #291916 |
When $n=1$ the two options for $h$ are $g_-$ and $g_+$, whose domains intersect only at $[\frac12,\frac12]$. If all of the domains at level $n$ intersect only at irrelevant points then at $n+1$ each $h_n$ becomes $h_n(g_-(x))$ and $h_n(g_+(x))$, which splits the domain of $h_n$ into two parts which o... (more) |
— | 6 months ago |
| Edit | Post #291919 | Initial revision | — | 6 months ago |
| Answer | — |
A: How can we prove that a point that follows another point has the same trajectory given a contant angle and ratio? The dot product cannot be sufficient because even making substitutions to minimise the occurrences of $Q$ we get $$\cos(\alpha)=\frac{\overrightarrow{P(t)}\cdot\overrightarrow{Q(t)}}{k|\overrightarrow{P(t)}|^2}$$ there are two solutions in the plane. I personally would tackle this by complex numbe... (more) |
— | 6 months ago |
| Edit | Post #291916 | Initial revision | — | 6 months ago |
| Answer | — |
A: Formalizing proof of existence of roots of functional equation All real roots are in $[0, 1)$. If $x \le 0$ then $f(x) \le x$ with equality only at $x = 0$. By induction, for all $n \in \mathbb{N}$ and $x \le 0$ we have $f^n(x) \le x$ with equality only at $x = 0$. If $x \ge 1$ then $f(x) \le 0$, so for all $n \in \mathbb{N}$ we have $f^n(x) \le 0 †... (more) |
— | 6 months ago |
| Comment | Post #291911 |
$f^n(x) - x$ is a polynomial of degree $2^n$, so it has $2^n$ roots. Is your goal essentially to prove that those roots are distinct real numbers in the interval $[0, 1)$? (more) |
— | 6 months ago |
| Comment | Post #290765 |
@#8046, yes, I did. (more) |
— | 6 months ago |
| Comment | Post #291827 |
It's not necessary to ask the same question three times, much less on the same day. (more) |
— | 6 months ago |
| Edit | Post #291828 | Question closed | — | 6 months ago |
| Comment | Post #290765 |
@#80779, physics has nothing to do with this discussion, but the question and the answer are both about philosophy of mathematics, so if your comments aren't philosophical then they're definitely missing the point. *Multiverse* here is a term taken from the paper which the question asks about. (more) |
— | 6 months ago |
| Comment | Post #290765 |
@xamidi, you're missing the larger point that this is a question of philosophy. You believe so completely in the multiverse philosophy that you're rejecting the possibility of existence of any other philosophy. (more) |
— | 6 months ago |
| Comment | Post #291691 |
It's a particular case of the [multiplicative group of integers modulo n](https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n). (more) |
— | 6 months ago |
| Edit | Post #291691 | Initial revision | — | 6 months ago |
| Answer | — |
A: All numbers are triangular modulo $N$ iff $N$ is a power of $2$? Only if: suppose $N = 2^a m$ with $m$ odd and greater than $1$. Because the odd numbers form a multiplicative group modulo $2^{a+1}$, there are $j0$, $j1$ such that $j0 m \equiv 1 \pmod{2^{a+1}}$ and $j1 m \equiv -1 \pmod{2^{a+1}}$. We have $(j0 + j1)m \equiv 0 \pmod{2^{a+1}}$ and since $m$ is odd... (more) |
— | 6 months ago |
| Comment | Post #291687 |
I think that your proof that odd $N$ doesn't work is the right idea for all non-powers of 2. The subgoal would be to show that if $N = 2^a m$ with $m$ odd and greater than $1$, there is a choice of (odd) $j < 2^a$ and sign such that $jm \pm 1 \equiv 0 \pmod{2^{a+1}}$; then according to the required c... (more) |
— | 6 months ago |
| Comment | Post #291660 |
You say "If someone doesn't want to adjust his answer" but how do we know that if they don't start the answer with "I don't want anyone to make even minor edits to this post"? The default assumption of this site is that people want to collaborate to build something together, not that people want to s... (more) |
— | 7 months ago |
| Comment | Post #291528 |
Your references appear to be missing the actual references. Which book or paper is "*Megginson*"? How about "*Theory of linear operations*"? (more) |
— | 7 months ago |
| Edit | Post #291473 | Initial revision | — | 7 months ago |
| Answer | — |
A: Would the following stats topics be in the scope of the math community? There is some overlap and it's not always easy to draw the line between mathematics and science (particularly theoretical physics), but most of the examples given seem to me to be more on the "using mathematics in other fields" side of the line than the "mathematics" side of the line. I think that... (more) |
— | 7 months ago |
| Comment | Post #291432 |
I find it very suspicious that you're performing an indefinite integral and getting a result with no constant of integration. Probably if you want to state the general formula with indefinite integrals you should include an explicit constant alongside $f(x)g(x)$. (more) |
— | 8 months ago |
| Comment | Post #291301 |
For the purposes of "each point in $X$ is in the image of exactly one such restriction $\sigma_\alpha | \overset{\circ}{\Delta}{}^n$" does the 0-simplex count as having a non-empty interior? If not, the points mapped by the vertex would seem to be exceptions. (more) |
— | 8 months ago |
| Comment | Post #291104 |
I haven't worked through it, but my first thought is that your diagram doesn't seem to use any properties of the circumcircle. I wonder whether the fact that the perpendicular bisector of CD is parallel to MD and passes through the centre of the circle will be relevant. (more) |
— | 9 months ago |
| Comment | Post #290814 |
The FAQ says "Don't cross-post the same thing in multiple topics". I believe that "topics" is a term which generalises "question" to cover sites which have blog posts, articles, etc. Cross-posting questions across sites / networks without mentioning it is poor netiquette, but here the OP did provide ... (more) |
— | 10 months ago |
| Comment | Post #290894 |
Yes, you're right. I don't think I've kept all of my previous Sage code so I can't try to figure out where I made the mistake, but that does open up some other ideas for extensions to $6 \times 6$. (more) |
— | 10 months ago |
| Comment | Post #290894 |
This can be rephrased as $$(A+I)^{-1} = \frac{(1 + a_{01}^2 + a_{02}^2 + a_{12}^2)I - A + A^2}{|A+I|}$$ where $|A+I| = 1 + a_{01}^2 + a_{02}^2 + a_{12}^2$. For $4\times 4$ and $5\times 5$ it generalises respectively to $$-\frac{(1 + a_{01}^2 + a_{02}^2 + a_{03}^2 + a_{12}^2 + a_{13}^2 + a_{23}^2)(A -... (more) |
— | 10 months ago |
| Comment | Post #290864 |
If you want to learn advanced mathematics, the best route is to work through a textbook. If you come to a proof that you can't understand, feel free to ask a specific question about that proof. Wikipedia mathematics pages are often more helpful as a refresher on something already studied than as a re... (more) |
— | 10 months ago |
| Edit | Post #290854 | Question closed | — | 10 months ago |
| Edit | Post #290864 | Question closed | — | 10 months ago |
| Edit | Post #290745 |
Post edited: Reduced the question metainformation to that which is relevant to this site |
— | 10 months ago |
| Edit | Post #290765 | Initial revision | — | 10 months ago |
| Answer | — |
A: Concrete examples of set theorists thinking independence proofs only determine provability rather than that a statement is neither true nor false? >> …the independence of a set-theoretic assertion from ZFC tells us little about whether it holds or not in the universe. > > How can this be? If it is independent, then it cannot be proved from the axioms. Thus, one has the freedom to assume it or assume the negation, as an axiom. Why would someon... (more) |
— | 10 months ago |
| Edit | Post #290642 |
Post edited: Fix typo |
— | 11 months ago |
| Edit | Post #290637 | Question closed | — | 11 months ago |
| Comment | Post #290581 |
The case analysis needs to take into account the possibility of a 2-2-0-0 split. If there are 4 voters and 4 candidates then there are $4^4 = 64$ ways for them to vote. (more) |
— | 11 months ago |