Wikipedia:Reference desk/Archives/Mathematics/2013 April 2

= April 2 =

For each every any all
I have some draft high school math textbook pages translated from Russian by a non-native (but quite proficient) English speaker. The definition of an even function is defined as "$f(-x) = f(x) for each x∈R$". The term "for each" is used in all such instances, as opposed to the usual "for all" (or possibly "for any").

There are two reasons why I'd ask them to change: First, "for any/all" is a cognate for the upside-down "A" symbol (∀). Second, the notion of "each" element of an infinite set seems to imply the use/necessity of the Axiom of Choice (and though ridiculously pedantic my friends think it's a valid objection). Is "for each" an acceptable term for an English text nonetheless? SamuelRiv (talk) 00:43, 2 April 2013 (UTC)
 * Your first reason is a good one; your second is baseless. You're not using the axiom of choice unless you make a choice for each element of the infinite set.
 * Another reason to avoid "for each" is because sometimes in English it reverses the order of the quantifiers. "There exists an x for each y" sounds like "for all, exists" instead of "exists, forall".--80.109.106.49 (talk) 02:03, 2 April 2013 (UTC)
 * I've seen it argued on this page (it seems, quite correctly) that "for any" should also be avoided in mathematical contexts due to similar ambiguity of the language. That seems to leave "far all" as the only reasonable alternative of those given, on purely linguistic grounds. — Quondum 09:33, 2 April 2013 (UTC)
 * I prefer 'for every' in that type circumstance. I think 'for all' is more common just I think it implies a property of the totality rather than of each member. Dmcq (talk) 10:03, 2 April 2013 (UTC)

"Time packing" Algorithm
Hi all,

At work we have a long list of tasks that need to be completed every quarter, each which have a designated "time to complete" and start/end dates. Currently these are allocated by hand to try to spread out the workload so that no person is given more than 7 hours work every day and everything is done on time, there are also some additional contraints such as some people only do one type of work.

I see this as a fairly simple packing algorithm with lots of constraints. Surely someone must have done some study into a similar situation - as it seems to arise in many workplaces, is there an algorithm that has been used by companies to do the same? Or does this situation have too many constraints to warrant an efficient algorithm?

Ideally I would like an algorithm that I can implement in a coding language to allocate work, given the constraints - is there such a thing? 80.254.147.164 (talk) 09:46, 2 April 2013 (UTC)


 * There's lots of project planning programs for doing this sort of thing, you're probably better off at the computer reference desk. Gantt charts and critical path are the sorts of words you'll see a lot of. Dmcq (talk) 10:15, 2 April 2013 (UTC)


 * Yes there is lots of work in this kind of thing, its a field worth billions of pounds. Its very similar to the way parallel processing works by scheduling tasks to different processors. Job shop scheduling looks very close to your problem.--Salix (talk): 12:22, 2 April 2013 (UTC)


 * I'd be cautious about allocating people like this, as the program isn't likely to account for human factors in the way a manager doing the same assignments might. For example, a manager might avoid assigning two people to the same task who don't get along.  In theory a computer program could do this, too, but the number of inputs are far more than you might think, such as the relationship of every worker with every other. StuRat (talk) 03:50, 4 April 2013 (UTC)

Sigma Additivity Definition Confusion
Are the following two definitions of sigma additivity equivalent? If so, how?
 * $$\mu\left(\bigcup_{i \in I} E_i\right) = \sum_{i \in I} \mu\left(E_i\right)$$ for all countable collections $$\left\{E_i\right\}_{i \in I}$$ of pairwise disjoint sets in any sample space $&Omega;$&thinsp;–&thinsp;i.e.: this statement $$\forall \left\{E_i\right\}_{i \in I} \iff I \subsetneq \mathbb{N} \land \bigcap_{i \in I} = \varnothing$$
 * $$ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n)$$ for any sequence of disjoint subsets $A_{ n }$ in any sigma algebra $$\mathcal{A}$$&thinsp;–&thinsp;i.e.: this statement $$\forall n$$

RandomDSdevel (talk) 20:13, 2 April 2013 (UTC)
 * I'm not totally sure I follow what the statements are, but I'll have a go. I believe these are inequivalent, but both are arguably incorrect definitions. To start with, by definition any countable set I is basically the naturals. This leaves us with two potential differences: 1) (pairwise) disjointness and 2) the sigma-algebra rather than being a general subset of $$\Omega$$. The disjointness is a problem because the word pairwise is often omitted from "disjoint" - the only other thing it could mean is that the whole intersection is empty, which in most cases I know of is clearly not what's wanted from context - both definitions are the same. For the second, if your function $$\mu$$ is defined on all subsets of $$\Omega$$, then this is the same thing with respect to the sigma-algebra "all subsets"; in general, however, it won't be (think of non-measurable sets). Does that help at all? Straightontillmorning (talk) 15:17, 4 April 2013 (UTC)