Activity for Peter Taylor
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Answer | — |
A: Do the Faber partition polynomials have integer coefficients? Let $B(x) = b1 x + b2 x^2 + \cdots$. Then $$\begin{eqnarray} Fn(b1, \ldots, bn) &=& - [x^n] n \log(1 + B(x)) \\\\ &=& [x^n] n \sum{i \ge 1} \frac{(-B(x))^i}{i} \\\\ &=& \sum{\lambda \\, \vdash \\, n} \frac{n}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f1, \ldots, fn} \prodj ... (more) |
— | almost 2 years ago |
| Edit | Post #286655 | Initial revision | — | almost 2 years ago |
| Question | — |
Do the Faber partition polynomials have integer coefficients? The Online Encyclopedia of Integer Sequences includes A263916: Coefficients of the Faber partition polynomials. Perhaps the clearest definition given is > -log(1 + b(1) x + b(2) x^2 + ...) = Sum{n>=1} F(n,b(1),...,b(n)) x^n/n which in better notation is $$\sum{n \ge 1} Fn(b1, \ldots, bn) \f... (more) |
— | almost 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) |
— | almost 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) |
— | almost 2 years ago |
| Edit | Post #286569 |
Post edited: |
— | almost 2 years ago |
| Edit | Post #286569 |
Post edited: Investigate with some simple cost functions |
— | almost 2 years ago |
| Edit | Post #286569 | Initial revision | — | almost 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) |
— | almost 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) |
— | almost 2 years ago |
| Edit | Post #286453 |
Post edited: Improve legibility |
— | almost 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) |
— | almost 2 years ago |
| Edit | Post #286399 | Initial revision | — | almost 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) |
— | almost 2 years ago |
| Edit | Post #286398 | Initial revision | — | almost 2 years ago |
| Question | — |
Matrices with rotational symmetry I've seen a claim without proof that the characteristic polynomials of matrices with rotational symmetry (i.e. $n \times n$ matrices $A$ with $A{i,j} = A{n+1-i,n+1-j}$) always factor into the product of the characteristic polynomials of smaller matrices which can be derived from blocks of the origina... (more) |
— | almost 2 years ago |
| Comment | Post #286034 |
@#53398, I'm not sure I get it either. If we have a rotation in a plane around the origin which, applied twice, takes us from $(1,0,0)$ to $(-1,0,0)$ then the plane must contain the origin, the start point, and the end point. By symmetry the square roots of $-1$ end up all being in the plane through ... (more) |
— | about 2 years ago |
| Comment | Post #286188 |
On the subject of having no idea how to answer a question, it's always a good starting point to write it out in terms of the definitions. So "the expected time until the buffer reaches its capacity for the first time" is a weighted sum of probabilities, and the question largely reduces to calculating... (more) |
— | about 2 years ago |
| Comment | Post #286188 |
This site uses a markup tool called MathJax which would improve the legibility of your question considerably. If you surround each formula/expression with dollar signs, you can use underscores `_` to create subscripts, carets `^` to create superscripts, curly brackets `{}` to group subscripts or supe... (more) |
— | about 2 years ago |
| Comment | Post #286166 |
Rather than thinking about going "out of 3D space", it may be helpful to think about it as working with the surface of a 3D unit hypersphere embedded in a 4D space. (more) |
— | about 2 years ago |
| Edit | Post #286130 |
Post edited: MathJaxify, trim some extraneous verbiage |
— | about 2 years ago |
| Edit | Post #286004 | Question closed | — | about 2 years ago |
| Comment | Post #286004 |
Which skewness and which sample skewness? There seem to be multiple, subtly different, parameters with those names. (more) |
— | about 2 years ago |
| Edit | Post #285984 |
Post edited: |
— | about 2 years ago |
| Comment | Post #285984 |
Quaternions are not complex numbers. The complex numbers can be seen as a subalgebra, but I'm not sure to what extent that is useful or helpful. (more) |
— | about 2 years ago |
| Edit | Post #285988 | Initial revision | — | about 2 years ago |
| Answer | — |
A: How to write the big Xi notation in MathJax? $\Xi$ may not be what you want, but (a) it's the correct answer to the question as stated; (b) I can't understand why you'd complain that the font used is more legible than a handwritten example deliberately chosen to be difficult to parse. $$\frac{\Xi}{\overline{\Xi}}$$ isn't marvellous, but it's ce... (more) |
— | about 2 years ago |
| Edit | Post #285682 | Initial revision | — | over 2 years ago |
| Answer | — |
A: How can a 15 year old construe the LHS of Generalized Vandermonde's Identity, when it lacks summation limits and a summation index? You can rewrite it in different notation with sum limits if you want. You just need to use a different way to express the constraint on which terms to include in the sum. E.g. with the Iverson bracket notation the LHS becomes $$\sum{k1 = 0}^m \sum{k2 = 0}^m \cdots \sum{kp = 0}^m [k1+\cdots +kp = m... (more) |
— | over 2 years ago |
| Edit | Post #285679 | Question closed | — | over 2 years ago |
| Edit | Post #285674 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Why 1. multiply the number of independent options? 2. add the number of exclusive options? You might find it more helpful to instead consider only two choices at a time and draw a grid: | beef | chicken | fish | ------+---------+---------+---------+ | beef | chicken | fish | fries | + | + | + | | fries | fries |... (more) |
— | over 2 years ago |
| Edit | Post #285672 | Question closed | — | over 2 years ago |
| Comment | Post #285671 |
Have you drawn a diagram of the squares in question? (more) |
— | over 2 years ago |
| Comment | Post #285015 |
@#55022, if you have a property which you do not find intuitive or counter-intuitive then you need a very good reason for believing that it *should be* intuitive to justify the question. Axioms and definitions are chosen and become popular because they're *useful*, with no guarantees about how intuit... (more) |
— | over 2 years ago |
| Edit | Post #285527 | Initial revision | — | over 2 years ago |
| Answer | — |
A: How's it possible to arrange 0 objects? How can 0! = 1? Combinatorics is not necessarily tangible, so the question of what it means to physically rearrange objects is irrelevant. On the other hand, it's easy to write down the permutations of small finite numbers of objects: 0: [] 1: [1] 2: [1,2] [2,1] 3: [1,2,3] [1,3,2] [2,1,3] [2,3,... (more) |
— | over 2 years ago |
| Edit | Post #285476 | Initial revision | — | over 2 years ago |
| Answer | — |
A: If C = Calvin wins the match, and $X \thicksim Bin(2, p) =$ how many of the first 2 games he wins — then why P(C|X = 1) = P(C)? The key is > the first player to win two games more than his opponent wins the match $P(C \mid X = 1)$ describes the situation when Calvin has won 1 of the first two matches: so the opponent won the other match, and they're currently even. Since a 1-1 score is equivalent to a 0-0 score for the ... (more) |
— | over 2 years ago |
| Edit | Post #285344 | Question closed | — | over 2 years ago |
| Comment | Post #285425 |
My impression (and I haven't just rechecked the question list to validate it) is that very few of the questions are at a level of needing a grad student to answer them. It's probably fair to say that those which *are* at that level are mostly unanswered, but it's also probably fair to say that that *... (more) |
— | over 2 years ago |
| Comment | Post #285342 |
Closed as off-topic because the issue at heart here isn't mathematics but linguistics. In particular, what a good answer would address is the flaws in assuming firstly that a word has only one meaning, and secondly that whoever made that YouTube video is the absolute authority on what that meaning is... (more) |
— | over 2 years ago |
| Edit | Post #285342 | Question closed | — | over 2 years ago |
| Comment | Post #285086 |
I don't think that MCD is "based on" either site, but my point is that its scope is similar to MSE's and so it's not a surprise that it has the same problems as MSE. (more) |
— | over 2 years ago |
| Edit | Post #285086 | Initial revision | — | over 2 years ago |
| Answer | — |
A: How can we grow this community? Certainly low quality content is an issue. Being brutally honest, if I understood the rôle of moderator as being a ruthless dictator who pursues quality above all else I would delete 95% of the questions. But to put that in context, it's also true for most maths fora. I recently saw someone mention a... (more) |
— | over 2 years ago |
| Comment | Post #284996 |
That graphic helps a lot, thanks. But if the two initial squares aren't horizontally or vertically adjacent, it's not clear that the border is well defined. You can draw the 8-square box around each one, and then connect appropriate corners of the boxes with straight lines, but deciding which squares... (more) |
— | over 2 years ago |
| Comment | Post #284996 |
The notation can be read "the set of points $(0, t)$ where the variable $t$ is between $0$ and $1$ inclusive". It's just the line segment (technically not a line, because lines are infinite) between the two points. (more) |
— | over 2 years ago |
| Comment | Post #284996 |
This seems to be getting less clear, not more. Take a concrete example: let the two points be $(0, 0)$ and $(0, 1)$. What is the "minimal border" if not the line segment $\\{(0, t) \mid 0 \le t \le 1\\}$? (more) |
— | over 2 years ago |
| Comment | Post #285015 |
Turn the question around: what is *unintuitive* about it? (more) |
— | over 2 years ago |