What exactly is an operation? I understand that multiplication and addition are operations, but what about the derivative ($\frac{\mathrm d}{\mathrm dx}$).

Can an operation be a relation of expressions, or just of numbers?

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    $\begingroup$ Yes, we usually think of differentiation as an operation! One way you can think about it is to take any polynomial expression (or something like a Taylor series of a non-polynomial expression) and re-express it as a (potentially-infinite) vector, where the value in the nth position is the coefficient on the $x^n$ term, starting at zero for the constant. Then you can express differentiation as this big (potentially-infinite) matrix, and differentiating this expression vector is just multiplying it by the matrix to produce a new vector. We always use the word “operator” for such things. $\endgroup$ – Jack Crawford Jun 7 at 0:30
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    $\begingroup$ Note: addition and multiplication are binary operations $\endgroup$ – J. W. Tanner Jun 7 at 0:42

It all depends on your definitions. Wikipedia defines an operation in the following way.

In mathematics, an operation is a calculation from zero or more input values (called operands) to an output value.

Since functions aren't really values, I wouldn't consider differentiation to be an operation, since it's an association between input functions and output functions, not input values and output values.

Operators are a generalization of the notion of an operation. They need not necessarily input or output values. More generally, the can input and output values of any prespecified set. According to Wikipedia, the definition of an operator is as follows.

In mathematics, an operator is generally a mapping that acts on elements of a space to produce elements of another space (possibly the same space, sometimes required to be the same space).

There are numerous sorts of classes of operators that have been given names in mathematics. One class that differentiation falls under is that of linear operators, which are operators acting on vector spaces which satisfy a particular property (you can check out more information here). In the context of differentiation, the key property that differentiation satisfies in order to be able to regard it as a linear operator is the following one, which holds for any functions $f,g$ and values $a,b$.

$$\frac{d}{dx}\left( af(x)+bg(x)\right) = a\frac{d}{dx}f(x) + b\frac{d}{dx}g(x)$$

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    $\begingroup$ If an operator can take maps values of any set to values from any other set, what is the difference between an operator and a relation? $\endgroup$ – Frasch Jun 7 at 0:47
  • $\begingroup$ @Frasch Briefly, an operator is a map whereas a relation is an element from one (or more than one) set (acted on by a map) to be mapped to the co-domain by the map. Quote: 'In mathematics, a binary relation over two sets A and B is a set of ordered pairs (a, b) consisting of elements a of A and elements b of B; in short, it is a subset of the Cartesian product A × B. It encodes the information of relation: an element a is related to an element b if and only if the pair (a, b) belongs to the set.' en.wikipedia.org/wiki/Binary_relation $\endgroup$ – Mathematicing Jun 7 at 0:50

In mathematics, an operation on the set $X$ is a function from a power of $X$ to $X$; that is, a function that takes some number of elements of $X$ and returns an element of $X$.

You can have finitary operations, that take a finite number of arguments. For example, the usual sum, difference, product of real numbers is an operation on the real numbers. The function that takes three $3$-dimensional vectors in $\mathbb{R}^3$, $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{z}$, and returns the vector $(\mathbf{u}\times\mathbf{v})\times\mathbf{z}$ is an operation on $\mathbb{R}^3$. And the function that takes a positive real number $x$ and sends it to $\frac{1}{x}$ is an operation on the positive real numbers.

A finitary operation has an arity, which is the number of arguments it takes. Sum, difference, product, are binary (or $2$-ary) operations. The one I described on $\mathbb{R}^3$ is a ternary (or $3$-ary) operation. The function $x\mapsto \frac{1}{x}$ on the positive reals is a unary (or $1$-ary) operation. (There are even nullary, or $0$-ary operations, but those are tricky, so forget about them for now).

There are also infinitary operations, that take infinitely many arguments, but let’s also ignore those for now.

Now, the first thing to note is that when you look at the differentiation “function”, $\frac{d}{dx}$, this takes as inputs functions and has functions as outputs. That’s a good step. But the next question is: what is our set?

Can’t be the set of all continuous functions, because not every continuous function has a derivative. It can’t be the set of differentiable functions, because the derivative of a differentiable function does not have to be differentiable, etc.

So, in the abstract, your question has no answer until we know what collection of functions you are thinking about.

Here are some collections of functions on which $\frac{d}{dx}$ is, indeed, a unary operation:

  1. The collection of all analytic functions $\mathbb{R}\to\mathbb{R}$; these are functions that have Taylor expansions around any point. The derivative of such a function is again such a function.

  2. The collection of infinitely differentiable functions. (slightly smaller than the collection of analytic functions).

  3. The collection of all polynomial functions (since the derivative of a polynomial function is a polynomial function).

  4. The collection of all functions expressible as polynomials in $\sin(x)$ and $\cos(x)$.

There are other such collections on which $\frac{d}{dx}$ is an operation. In fact, many such collections form a vector space, and $\frac{d}{dx}$ will be a linear operator (a linear transformation from the vector space to itself).

For other sets, such as the collection of all differentiable real valued functions of real variable, you technically don’t have an operation, because the image of such a function may fall “outside” the initial set.

  • $\begingroup$ I am reminded of your other, more famous answer... speaking of which, by "slightly smaller than the collection of analytic functions", did you have things like $\exp(-x^{-2})$ in mind? $\endgroup$ – J. M. is a poor mathematician Aug 15 at 5:21
  • $\begingroup$ @J.M.isapoormathematician: I wasn’t thinking of any particular example (I just know that there are functions that are smooth but not analytic). A common example is the function that is $0$ for $x\leq 0$, and $\exp(-\frac{1}{x})$ for $x\gt 0$. $\endgroup$ – Arturo Magidin Aug 15 at 5:53

$\frac{d}{dx}$ is a linear operator.

And expanding on the comment made by Jack Crawford, a linear operator has a matrix representation the image of which is completely dependent on the choice of basic set of the finite vector space on which it is acting on.

  • $\begingroup$ That’s only true when the vector space is finite dimensional (unless you extend the notion of “matrix representation” to “infinite matrices”). In addition, it is wrong to talk about the basis of a vector space, as most vector spaces have many bases (in fact, the only ones that have a unique basis are the zero dimensional space over any field, and a one dimensional space over the field of 2 elements). $\endgroup$ – Arturo Magidin Jun 7 at 1:01
  • $\begingroup$ @ArturoMagidin Should have worded it better - good catch. $\endgroup$ – Mathematicing Jun 7 at 1:06

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