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April 11

Group (one op), rings (two ops), ???? (three ops)

What would be the three operation analogue of groups and rings? More than three operations? I'm fully aware of things like algebras that have three ops but that doesn't seems as general as it should be. The closest term I've found is the generic Algebraic structure. Is there no widespread convention in use about what to name or how to definition, higher operation alegebric structures on a single set? Jason Quinn (talk) 01:31, 11 April 2014 (UTC)[reply]

A group isn't any structure with one op, it is a structure that satisfies specific constraints; the same for rings in the case of two ops. In short, groups are not generic one operation structures - nor are they the "assumed" structure with one operation. The main reason that we tend to study groups/rings/etc. is because they behave nicely, for the most part, and are useful for our interests; but, they are neither more plentiful, more natural, nor more "appropriate" than any other set of axioms and functions attached to a set.Phoenixia1177 (talk) 08:32, 11 April 2014 (UTC)[reply]

The wisdom may be that with too few axioms, then nothing interesting can be said. With too many, then there are very few examples. The "usual" structures are probably the one that strikes the right balance. YohanN7 (talk) 15:49, 11 April 2014 (UTC)[reply]

My question could perhaps be re-phrased as, "Is there a 'usual' structure that mathematicians have agreed upon for more than two operations that 'strikes the right balance' in the same way that groups and rings do?" The answer appears to be 'no'. The classical sets like Z, Q, and R are the key objects from which the ideas of groups and rings were distilled. In the absence of a third operation of equal importance to normal + and *, maybe there's just not motivation to study higher operational objects as intensely. It's also possible that there's a deeper reason of which I am unaware that negates the need for further classification. For instance, if one were to simply require the new operations be associative and obey the distributive law like the first multiplicative operation, then clearly, the set is some sort of "multi-ring". If people have studied "multi-rings" perhaps it just turns out they are uninteresting and without much interplay between the operations. I don't know. Jason Quinn (talk) 00:17, 12 April 2014 (UTC)[reply]

Somebody who knows should really answer, but anyway, here are my 2c:

I think your analysis is correct except for the last point. I'd not think "Multi-rings" would be uninteresting, there are just too few interesting examples. On the other hand, you have structures that have plenty of operations, like a graded tensor algebra. Combine (tensor) one of those with its dual, and you get even more. You can also drop the good-old associativity requirement, like in Lie algebras. There are plenty of those, and they are certainly interesting. They may, however, not be as clean-cut as you seem to be looking for. YohanN7 (talk) 11:32, 12 April 2014 (UTC)[reply]

A set with one closed binary operation is called a magma. I doubt whether there is any use for a generic category with two such operations, and even less so with three. —Quondum 17:17, 11 April 2014 (UTC)[reply]

CALCULUS question: Formula for integral of tan x

According to one online source, the formula for the integral of tan x is -ln (cos x) + c. According to my math book, the formula is ln (sec x) + c

(by the way, I am aware how this formula is derived by changing the problem to integral of (sin/cos) and using u substitution)

I'm taking a test tomorrow and if I had to integrate tan x for one of my problems, would my professor (assuming he is competent and fair, which he seems to be) accept either answer?

This raises the larger question, if I'm not mistaken, of logarthmic properties. sec and cos, of course, are reciprocals. So does the natural log of x ALWAYS equal the negative natural log of (1/x). When I asked wolframalpha.com this question, it said no, this proposition is false.

I believe this is because when x is negative, it doesn't really hold true (or maybe it does???). Also, since cosines and secants can sometimes be negative, maybe this causes a problem such that I have to be careful whether I say ln (sec x) or -ln (cos x). Can someone help? --24.228.94.244 (talk) 21:45, 11 April 2014 (UTC)[reply]

They have the same value. ln(sec x) = ln(1/cosx) = -ln(cos x) (note you want to keep cos x positive to make the ln work. Try putting in ln(sec x) + ln(cos x) where -(pi/2) < x < (pi/2) into wolfram alpha. While the graph looks jagged, notice that the y values are on the order of magnitude of 1x10^-16, which means that it is dealing with rounding errors, but the value is more or less zero.Naraht (talk) 21:55, 11 April 2014 (UTC)[reply]

Your "larger question" is true, because (for any base, not just natural log) log (x^n) = n log (x). I can't think what the problem was with Wolfram. And yes, your professor would be able to work all of this out, since it is his bread and butter. And as an occasional teacher/tutor, you really should revise that stuff about logs and exponentials, since it is very important. IBE (talk) 00:35, 14 April 2014 (UTC)[reply]

The problem that Wolfram Alpha may see could be that ln(1/x)=-ln(x) is no longer unconditionally true if you consider complex numbers. Whether it is true or not will depend on your choice of branch cut for the complex logarithm function. For the most common branch cut along the negative real axis (which gives rise to the principal logarithm function Log z), it will be true for any complex number that is not a negative real (or zero, of course). If you cut along the positive axis and define the complex logarithm such that it extends the real logarithm, the statement log(1/x)=-log(x) will be true only for positive real numbers x. —Kusma (t·c) 15:05, 15 April 2014 (UTC)[reply]