View Full Version : Operators and Relations
Swapnil
Oct16-06, 03:04 PM
I was wondering, what is the difference between an operator and a relation? For example, instead of saying 2+3 I can say Add(2,3). Or the \frac{df(x)}{dx} operator can be written as D(f(x)).
I fail to see any difference between an operator and a relation. What do you guys think?
Do you know the definitions of "operator" and "relation"?
HallsofIvy
Oct16-06, 03:58 PM
I don't see where any of those examples have to do with "relation".
Yes, 2+ 3 could be called Add(2,3) but neither of those is a relation.
Yes, \frac{df(x)}{dx} can also be written D(f(a)) but both of those are operators.
A "relation" is a set of ordered pairs. I don't see any "relations" in what you have written.
Swapnil
Oct17-06, 08:52 PM
I am sorry, but I was trying not to use the term "function" to avoid ambiguity and ended up using the term "relation" to make my question even more illogical.
I am trying to use the term "function" is a loose sense. The sense where you think of it as a black box which spits out an output given an input. I don't know if there is a mathematical name for this entity...
So what I meant was that you can think of the an operator as the same entity, where you have a bunch of inputs and the operator combines the input in a specific way and spits out an output.
I am still podering about this question and I am sorry if my question sounds ambigious or silly.
I am sorry, but I was trying not to use the term "function" to avoid ambiguity and ended up using the term "relation" to make my question even more illogical.
I am trying to use the term "function" is a loose sense. The sense where you think of it as a black box which spits out an output given an input. I don't know if there is a mathematical name for this entity...
So what I meant was that you can think of the an operator as the same entity, where you have a bunch of inputs and the operator combines the input in a specific way and spits out an output.
I am still podering about this question and I am sorry if my question sounds ambigious or silly.
I would say you have done a step forward in the path to abstraction. Of course, the sum and the derivative are functions (or functionals). The only thing of their "strange" notation is that they are so common, and this strange notation has been used for so long, that it makes no sense to "standarize" their notation. Also, the notation is useful (less characters to type).
HallsofIvy
Oct18-06, 06:16 AM
In common mathematical parlance, an "operator" is a function that is applied to functions. That is, the basic definition of "function" is that a function is a set of ordered pairs (a "relation") such that no two pairs have the same first member. There is nothing in that that says the members of the ordered pairs have to be numbers. In that sense, we can think of the derivative as a function that contains such ordered pairs as (x2, 2x), (sin x, cos x), and (ex, ex). Because the members of the ordered pairs are functions rather than numbers that is considered an "operator".
In common mathematical parlance, an "operator" is a function that is applied to functions. That is, the basic definition of "function" is that a function is a set of ordered pairs (a "relation") such that no two pairs have the same first member. There is nothing in that that says the members of the ordered pairs have to be numbers. In that sense, we can think of the derivative as a function that contains such ordered pairs as (x2, 2x), (sin x, cos x), and (ex, ex). Because the members of the ordered pairs are functions rather than numbers that is considered an "operator".
HallsofIvy, I do not know if this is purely "naming", but I would say that a "function" that is applied to "functions" is a functional (maybe this is more used in physics, though). Instead, an "operator" is something that may describe a "function" or a "functional". In functional analysis, for example, theorems are for operators, and they may apply either to spaces of finite dimension ("functions") to to spaces of infinite dimension ("functionals").
HallsofIvy
Oct18-06, 10:22 AM
I wouldn't argue the point! I don't think "operator" is as precisely defined in mathematics as "function", "relation", or even "functional".
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.