# Wikipedia:Reference desk/Archives/Mathematics/2011 April 12

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# April 12

## Integral formulas

Is there any formulas for indefinite integrals (antiderivative)? —Preceding unsigned comment added by 123.24.70.186 (talk) 04:54, 12 April 2011 (UTC)

Yes. Many. See, for example, Lists of integrals. —Bkell (talk) 05:06, 12 April 2011 (UTC)
... but there is no formula or method that will return the antiderivative of any given function. Methods that are known, such as the Risch algorithm, only work for certain classes of functions. Gandalf61 (talk) 13:17, 12 April 2011 (UTC)
Well, in one sense the fundamental theorem of calculus could be said to be such a formula. Michael Hardy (talk) 15:10, 12 April 2011 (UTC)
More of an existence theorem, surely ? Gandalf61 (talk) 15:31, 12 April 2011 (UTC)
Nonelementary integral might be what you're looking for. Dmcq (talk) 16:15, 12 April 2011 (UTC)

## Graph factorization

Resolved

Given that a nonempty graph G is 2 factorable if and only if G is 2n regular for some n>0, I wish to show that a 2n regular graph has a 2m factor for each m<n. Moreover why is the complete graph on even vertices always 1-factorable. Thanks-Shahab (talk) 06:01, 12 April 2011 (UTC)

A 2n-regular graph is 2-factorable, so it can be factored into n edge-disjoint 2-factors. Take the union of any m of those 2-factors. For your second question see Graph factorization#Complete graphs. —Bkell (talk) 06:39, 12 April 2011 (UTC)
How can we say that as a 2n-regular graph is 2-factorable, so it can be factored into exactly n edge-disjoint 2-factors? Why not more or less edge-disjoint 2-factors?-Shahab (talk) 06:56, 12 April 2011 (UTC)
A 2-factor is a spanning 2-regular subgraph, i.e., a subset F of the edges such that every vertex in the graph is an endpoint of exactly 2 edges in F. If the graph is 2n-regular, and the 2-factors are edge-disjoint, there must be n of them. (Look at any single vertex and consider the 2n edges incident upon it; every 2-factor uses up exactly 2 of them.) —Bkell (talk) 07:08, 12 April 2011 (UTC)
Thanks Bkell :) -Shahab (talk) 07:22, 12 April 2011 (UTC)

## Latex problem

When you talk about the 'latex file that I created on a Linux OS', I'm assuming you mean the .tex files rather than, say, a pdf or dvi file. Often compatibility problems between different operating systems like this are caused by different line endings being used. Most text editors are capable of converting between these. Apparently Protext includes TexnicCenter, which I've used before; as far as I can remember there is an option in one of the menus to select Windows/Linux/Mac-style line endings. Making sure that all of the input files are in Windows format might solve the problem (you might also need to delete any auxiliary files created by latex, if you copied them across from linux). 130.88.134.131 (talk) 11:25, 12 April 2011 (UTC)
Also, this question is probably more appropriate to the Computing refdesk: you might have got faster/better answers there. 130.88.134.131 (talk) 11:29, 12 April 2011 (UTC)

## Ambigious or not?

48÷2(9+3)=? The answer could be 2 or 288 depending on the order in which the "subroutines" are done, but which one is correct? Roger (talk) 17:40, 12 April 2011 (UTC)

288, as multiplication and division processes are done in the written order, so after evaluating the brackets, 48÷2*12=24*12=288. Grandiose (me, talk, contribs) 17:59, 12 April 2011 (UTC)
Yes, order of operations says the answer is 288. Invrnc (talk) 18:00, 12 April 2011 (UTC)
However, there is some ambiguity among humans in doing this operation; if we'd been doing 48÷2(12), a lot of people I think would get 2. Grandiose (me, talk, contribs) 18:03, 12 April 2011 (UTC)
The confusing part is the parens. You definitely need to do the part in parens first, but some people might mistakenly do the multiplication by the value in parens next. More parentheses is the way to avoid this problem: (48÷2)(9+3) or 48÷(2(9+3)). StuRat (talk) 18:55, 12 April 2011 (UTC)
I think the confusing part is the lack of parentheses... --COVIZAPIBETEFOKY (talk) 19:14, 12 April 2011 (UTC)
(EC) It does come down to Operator associativity, multiplication and division have the same precedence level and are left-associative. So you do the operations in written order as Grandiose says. All programming languages treat +, - , * and / as left associative. The associativity of operations tend to be glossed over in school and the following the BODMAS rule literally could lead to problems. Consider 7 - 2 + 3, note that addition comes before subtraction so you might interpret is 7 - (2 + 3) rather than using left-association (7 - 2) + 3.--Salix (talk): 19:23, 12 April 2011 (UTC)
Note that the J (programming language) interpretes 7-2+3 as 7-(2+3) and evaluates it to 2, while (7-2)+3 evaluates to 8. Bo Jacoby (talk) 07:42, 13 April 2011 (UTC).
I feel I should point out that the notion of a "left associative" binary operator is almost entirely limited to computer programming languages and grade-school arithmetic. In real mathematics, binary operators are associative or nonassociative, and nonassociative operators are always parenthesized (with rare exceptions like function application in the lambda calculus). The expression x − y + z is a shorthand for x + (−y) + z, where the only binary operator is +, which is associative. The division slash (mathematicians never use ÷) is not a binary operator, but a typographical variant of the horizontal division line. A mathematician who wrote $48 / 2 (9 + 3)\!$ might mean either $\frac{48}{2(9+3)}$ or $\frac{48}{2}(9+3)$, but probably the former. -- BenRG (talk) 05:44, 14 April 2011 (UTC)

It's fairly tedious. The basic primative recursive functions are quite easy to define in PA, such as addition, and 0, and projections. Compositions are fairly easy too. The hard part, as you'd imagine, is showing how to do primitive recursion. The trick here is to be able to pairing. Godel first did by proving Godel Beta function lemma which gives a $\Delta_0$ way to pair in PA. 69.95.83.137 (talk) 00:14, 13 April 2011 (UTC)