Activity for Derek Elkins
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #280068 |
Perhaps ironically, for CPUs, computing $1/\sqrt x$ is often faster than computing $\sqrt x$. One way to see why this might be so is to compare the Newton-Raphson iterations. For $y=\sqrt x$, we get: $y_{n+1}=y_n/2+x/(2y_n)$. For $y=1/\sqrt x$, we get: $y_{n+1}=y_n(3-xy_n^2)/2$. The key difference he... (more) |
— | over 3 years ago |
| Edit | Post #280133 | Initial revision | — | over 3 years ago |
| Question | — |
Marketing Math Codidact I don't think it's controversial to say that this site is currently suffering from a lack of new questions. I believe the main problem is lack of awareness. People aren't coming to this site because they have no idea it exists. A distant second concern, in my opinion, is the value proposition rela... (more) |
— | over 3 years ago |
| Comment | Post #280120 |
I agree that this is a fairly clumsily presented definition. A precise and reasonably easy to understand definition is to define it inductively with the base case, $\prod_{\nu=1}^0 x_\nu = e$, and the inductive case: $\prod_{\nu=1}^{n+1} x_\nu = \left(\prod_{\nu=1}^n x_\nu\right)\cdot x_{n+1}$. (more) |
— | over 3 years ago |
| Comment | Post #280118 |
Assuming $x_{(-)} : J \to G$, i.e. $x$ is a $J$-indexed collection of elements of $G$, then it can't be the case that $J \neq \varnothing$ while $\\{x_\nu \in G \mid \nu \in J\\} = \varnothing$. You could define a notion of product that took a finite multiset and compare that to a set-indexed notion ... (more) |
— | over 3 years ago |
| Edit | Post #280069 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why always rationalize a denominator? This answer combines the thoughts in both of the other (current) answers (tommi's and r's) but presents it in a more formal context. A significant portion of high school (and earlier... and a decent amount of later...) math is highly algorithmic, and, in particular, corresponds to the important id... (more) |
— | over 3 years ago |
| Edit | Post #279055 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why does “unless” mean “if not”? The quotes from the books are exactly the kind of ridiculously naive and over-simplified linguistics in introductions to logic that I criticize in this blog article. Particularly for everyday natural language expressions, you will find no shortage of ways where this "translation" provides only a clum... (more) |
— | over 3 years ago |
| Edit | Post #278447 |
Post edited: |
— | over 3 years ago |
| Edit | Post #278447 |
Post edited: |
— | over 3 years ago |
| Edit | Post #278447 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Existence of a set of all sets Sure. You can simply add $\exists V.\forall x.x \in V$ as an axiom to ZF(C), and you will have such an axiomatic system. Such an addition to ZF(C) would make it inconsistent, but it would still prove the existence of a "set of all sets" (along with everything else). As Peter Taylor points out in a... (more) |
— | over 3 years ago |
| Comment | Post #278382 |
Your first sentence is also potentially misleading. Math Codidact is also for people who aren't studying math and aren't professionals in related fields. We don't care (or have anyway of knowing) who the asker is or what their motivations are as long as the question is good and on-topic. (more) |
— | over 3 years ago |
| Comment | Post #278382 |
This is useless. People don't need detailed guidance about topics that are obviously on-topic anymore than they need guidance about topics that are obviously off-topic. This fails to accomplish even that as it's incomplete, as you say, and, more importantly, topicality isn't just a matter of subject ... (more) |
— | over 3 years ago |
| Edit | Post #278269 | Post edited | — | over 3 years ago |
| Comment | Post #278270 |
A problem that is almost certainly related is that sometimes the Markdown processor will convert substrings delimited by underscores to italics when those underscores are part of the MathJax, thereby breaking the MathJax. Peter Taylor's issue and this one are almost certainly due to the Markdown proc... (more) |
— | over 3 years ago |
| Edit | Post #278280 | Initial revision | — | over 3 years ago |
| Question | — |
Can we constructively find a third element of a set $X$ satisfying $X \cong X \times X$ given two distinct elements? Consider a set $X$ such that $X \cong X\times X$. It's quite easy to prove in a classical set theory, e.g. ZF, that $X$ must be the empty set or a singleton set or an infinite set. In other words, if we additionally assume that there exists $a, b \in X$ and $a \neq b$, then we know that $X$ is an inf... (more) |
— | over 3 years ago |
| Comment | Post #278270 |
I've suggested an edit that "fixes" the post (so if the edit is accepted people should look in the history to see the original form). That said, the intuition of the change I made is that the backslashes needed to be escaped. However, they probably shouldn't need to be escaped and so some MathJax or ... (more) |
— | over 3 years ago |
| Suggested Edit | Post #278269 |
Suggested edit: There seems to be some issue escaping backslashes(?). This edit fixes the issue now but may cause problems (extra vertical space) if the underlying issue is resolved. (more) |
helpful | over 3 years ago |
| Edit | Post #278145 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Computational hardness of the uniform halting problem According to Wikipedia, the set of (indices of) Turing machines that compute total functions, i.e. which halt on all inputs, is a $\Pi2$ set. If we use the variation of the definition of arithmetical hierarchy which includes primitive recursive functions, then it is fairly straightforward^[If you are... (more) |
— | over 3 years ago |
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