Activity for The Amplitwist
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #280206 |
+1 This makes perfect sense. If this is what the author also meant, then I'm not sure why he phrased it as $\mathcal{M}(k) = k$, though. (more) |
— | over 3 years ago |
| Edit | Post #280204 | Initial revision | — | over 3 years ago |
| Question | — |
Given a triangle with squares on two sides, the line segments joining the centres of the squares to the midpoint of the third side are equal and perpendicular I am reading Tristan Needham's Visual Complex Analysis (2012 reprint, OUP), and in $\S$1.III.3: Geometry, the author gives a geometric proof of the following fact, shown in Figure [12b] on page 16: >![Figure[12a] shows an arbitrary quadrilateral with squares constructed outward on each of the side... (more) |
— | over 3 years ago |
| Comment | Post #278724 |
This is happening to me, too. The editor slows down tremendously and drafting moderately long posts becomes quite difficult. (more) |
— | over 3 years ago |
| Edit | Post #280172 | Initial revision | — | over 3 years ago |
| Answer | — |
A: How can I deduce which operation removes redundacies? Let's take a look at the computation once more. The number of ways of arranging $3$ people in a line is $3! = 3 \cdot 2 \cdot 1 = 6$. That is, for any choice of $3$ people from the group of $8$ people, there are $3!$ ways to arrange them in a line. That is, there are $3!$ redundancies per choice o... (more) |
— | over 3 years ago |
| Comment | Post #280118 |
@MonicaCellio I tried editing the tags on the post just now and was somehow able to create new tags! I'm not sure what went wrong for me earlier. Thank you for pinging, though. :-) (more) |
— | over 3 years ago |
| Edit | Post #280118 |
Post edited: edited tags |
— | over 3 years ago |
| Comment | Post #280118 |
@PeterTaylor Yes, that's right. I included more than just that part in the quote because I felt that it provided context for what Lang is trying to convey; hopefully, it isn't too confusing! (more) |
— | over 3 years ago |
| Comment | Post #280134 |
@celtshk That actually sounds quite feasible, I'll support that! (more) |
— | over 3 years ago |
| Edit | Post #280134 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Marketing Math Codidact One long-standing issue on Math StackExchange is the policy on "problem-statement questions", or PSQs in short. The current policy there is that questions must provide sufficient context, and there is a close reason specifically for questions that lack context. There are highly active users on Math S... (more) |
— | over 3 years ago |
| Comment | Post #280118 |
@DerekElkins In my mind, I was thinking of a finite monoid (or group) $G$ whose elements are indexed as $\\{ x_1, \dotsc, x_n \\}$ and $I = \mathbb{N}$, $J = \\{ n + 1\\}$. But perhaps the notion of indexing already assumes a bijection, so taking $I = \mathbb{N}$ is improper? (more) |
— | over 3 years ago |
| Comment | Post #280118 |
I would have preferred to have some algebra-related tags on my question, but it seems that such tags don't already exist and I cannot create them. Could users with more privileges please help me choose a better set of tags? (more) |
— | over 3 years ago |
| Comment | Post #280064 |
A similar question asked on Mathematics Educators SE (sharing because there are some interesting answers there too): [How to justify teaching students to rationalize denominators?](https://matheducators.stackexchange.com/q/1860) (more) |
— | over 3 years ago |
| Edit | Post #280118 |
Post edited: fixed braces that weren't appearing |
— | over 3 years ago |
| Edit | Post #280118 | Initial revision | — | over 3 years ago |
| Question | — |
Product of empty set of elements vs. product over empty indexing set — is there any difference? I am reading Lang's Algebra (3rd ed., Pearson, 2003). In $\S$I.1 Monoids, on page 4 the author defines the meaning of and notations for products of finitely many elements of a monoid as follows: > Let $G$ be a monoid, and $x1, \dotsc, xn$ elements of $G$ (where $n$ is an integer $> 1$). We define... (more) |
— | over 3 years ago |
| Comment | Post #280069 |
This is a really insightful answer! +1 (more) |
— | over 3 years ago |
| Edit | Post #278629 |
Post edited: |
— | over 3 years ago |
| Comment | Post #278638 |
But, $x_{r-j}$ is *not* equal to $n^j x_r$! (I presume you meant modulo $10$?) For instance, for the decimal expansion of $1/19$, $x_{r-5} = 3$ but $2^5 x_r = 32 \neq 3 \pmod{10}$. In fact, they are not equal precisely because of the carrying-over happening in the concatenation process. I'm sorry, bu... (more) |
— | over 3 years ago |
| Comment | Post #278638 |
Secondly, though I say this "works", how do I adequately make sense of that? After all, the string that I get by my method of concatenation strictly speaking gives an "infinite number" such as $\dotsc 052631578947368421$. So perhaps, my claim is something like $$\frac{1}{10n-1} = \lim_{k \to \infty} ... (more) |
— | over 3 years ago |
| Comment | Post #278638 |
Thank you for your answer! I hope you won't mind if I say that I somehow feel this answer doesn't go all the way towards explaining why this "*backwards* concatenation" process appears. Specifically, I do understand that if $x = 0.\overline{x_1 x_2 x_3 \dotso x_r}$, then $x$ is a rational number equa... (more) |
— | over 3 years ago |
| Edit | Post #278629 |
Post edited: |
— | over 3 years ago |
| Edit | Post #278629 | Initial revision | — | over 3 years ago |
| Question | — |
Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern? I was playing around with the reciprocals of some positive integers and found these interesting patterns: $$ \frac{1}{19} = 0.\overline{052631578947368421} $$ Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows: \begin{a... (more) |
— | over 3 years ago |
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