April 12

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

Yes. Many. See, for example, Lists of integrals. —Bkell (talk) 05:06, 12 April 2011 (UTC)[reply]

... 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)[reply]

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)[reply]

More of an existence theorem, surely ? Gandalf61 (talk) 15:31, 12 April 2011 (UTC)[reply]

Nonelementary integral might be what you're looking for. Dmcq (talk) 16:15, 12 April 2011 (UTC)[reply]

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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)[reply]

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)[reply]

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)[reply]

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)[reply]

Thanks Bkell :) -Shahab (talk) 07:22, 12 April 2011 (UTC)[reply]

I'm having a lot of problems with latex. I'm using windows xp so I downloaded the following version on http://www.tug.org/protext/ from the link "download the self-extracting protext.exe file". I installed every package (it's 1.66gb and somehow took more than an hour to install). Now I had a latex file that I created on a Linux OS using a different version of latex, but for the windows version (exe called TeXworks) I could only create a pdf using the pdfLaTeX+MakeIndex+BibTeX option at the top of the screen; options such as pdfTeX and pdfLaTeX and XeTeX and XeLaTeX give errors. Furthermore, the "pdf" I created using pdfLaTeX+MakeIndex+BibTeX isn't really a pdf file; it's not opened in adobe pdf and there's no save option. It's opened in TeXworks, but looks like pdf files on the inside. I tried reading the "help" but I have no idea what they're on about. Can someone tell me in layman words how to use the windows version? Does the windows version read the linux version properly? Thanks in advance. There's a lot less options in the windows version than the linux version. Money is tight (talk) 09:17, 12 April 2011 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Yes, order of operations says the answer is 288. Invrnc (talk) 18:00, 12 April 2011 (UTC)[reply]

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)[reply]

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)[reply]

I think the confusing part is the lack of parentheses... --COVIZAPIBETEFOKY (talk) 19:14, 12 April 2011 (UTC)[reply]

(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)[reply]

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).[reply]

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 ${\displaystyle 48/2(9+3)\!}$ might mean either ${\displaystyle {\frac {48}{2(9+3)))}$ or ${\displaystyle {\frac {48}{2))(9+3)}$, but probably the former. -- BenRG (talk) 05:44, 14 April 2011 (UTC)[reply]

I begin by quoting a statement off of the Primitive recursive arithmetic page:

"In first-order arithmetic, the only primitive recursive functions that need to be explicitly axiomatized are addition and multiplication. All other primitive recursive predicates can be defined using these two primitive recursive functions and quantification over all natural numbers. Defining primitive recursive functions in this manner is not possible in PRA, because it lacks quantifiers."

My question is, how exactly does one go about defining a primitive recursive function (other than addition or multiplication) in Peano arithmetic, "using these two primitive recursive functions and quantification over all natural numbers," as opposed to a recursive definition? For example, what is the typical non-recursive definition of the factorial function in Peano arithmetic?
Thanks. JamesMazur22 (talk) 22:03, 12 April 2011 (UTC)[reply]

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 ${\displaystyle \Delta _{0))$ way to pair in PA. 69.95.83.137 (talk) 00:14, 13 April 2011 (UTC)[reply]

Thanks; I'll research that.

And now I understand. Any theorem involving primitive recursive functions/predicates can be proved using only addition and multiplication. And that fact is what is not true of primitive recursive arithmetic. JamesMazur22 (talk) 16:52, 18 April 2011 (UTC)[reply]

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