It's quite common nowadays to name a function and the application of the function to its input with the same letter. (Possibly more so in applied areas. Certainly many calculus textbooks do this.)

When did this practice start?

In particular, did any of the old masters like Newton, Leibniz, Euler etc. ever write something like $y=y(x)$?

Clarification: The question is really about the history of this practice. With whom did it start? I didn't want to discuss merits or demerits of this notation. If you want to provide a non standard interpretation of $y=y(x)$, please also back it up with historical references.

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    $\begingroup$ if you understand = as the operator of assignment (rather than equality), there is not really any "abuse of notation" --- $y=y(x)$ then assigns $y$ to be a function of $x$. $\endgroup$ Commented Oct 24, 2016 at 9:58
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    $\begingroup$ @CarloBeenakker But I have to add, I don't understand your interpretation at all. The $y$ on which side is being assigned to what? And before you made that assignment, what was the meaning of $y$ (a variable or a function?) $\endgroup$ Commented Oct 24, 2016 at 10:23
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    $\begingroup$ math.stackexchange.com/questions/636332/… doesn't answer the question, but may be worth a look. $\endgroup$ Commented Oct 24, 2016 at 11:42
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    $\begingroup$ My personal conjecture (after browsing some texts) is that we cannot find it in the "founding fathers" : Newton, Leibniz, Euler, etc. Th possible source can be the Lagrange-Cauchy notation for derivative; when $y=f(x)$, Lagrange uses indifferently $f'(x)$ as well as $y'$. Maybe, the source for the sloppy notation was the first that "commented" a cartesian diagram of speed vs time with the formula : $v=v(t)$... $\endgroup$ Commented Oct 26, 2016 at 10:06
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    $\begingroup$ This would be more on topic at hsm.stackexchange.com. $\endgroup$
    – user21349
    Commented Nov 1, 2016 at 15:35

5 Answers 5


Regarding the original question of who started literally writing $y=y(x)$ or something like it, which I understand Jacobi didn't do in the quoted 1840 paper: Cayley (1859, p. 3),

where $\Omega$ is regarded as a function of $r,v,y,$ or (as this may be expressed) where $\Omega = \Omega(r, v, y)$

sounds like an early example, in that he feels the need to explain the notation.


This seems to have started with Jacobi around 1840, when he re-introduced the popular notation for partial derivatives in De determinantibus functionalibus. He even provides a well-intentioned justification for starting this abuse of notation: he wanted a less ambiguous notation.

While reading Jacobi one should bear in mind, that during his time the $f$ in $f(x)$ was not yet officially called the function. This modern use only started after 1900 and after Frege, Dedekind, Peano and Cantor. (For more on this see Who first considered the $f$ in $f(x)$ as an object in itself, and who decided to call it a function?.) Prior to 1900 what was called the function was the $f(x)$ or the $y$ in $y=f(x)$, in other words the variable quantity, and not the "rule" $f$. This is also how Jacobi uses the word function below. Beware hence, that what he denotes with $f$ and calls a function, is actually a variable quantity. (In modern terminology: his $f$ is an object of type $\mathbb{R}$ and not of type $\mathbb{R}\to\mathbb{R}$.)

What follows is my translation of the German translation Ueber die Functionaldeterminante, 1941 p.2 ff:

Before I come to the actual subject matter, I will start with some remarks concerning the notation of partial differentials. And since this treatise will contain repeated talk about functions, which may or may not depend from one another, it seems appropriate to also add some elementary considerations about these.

$$2. $$

To distinguish the partial differentials from the ordinary ones, hence from those where all variable quantities are seen as functions of a single variabel, Euler and others put the partial differentials in between brackets. But since an accumulation of brackets becomes rather annoying for reading and writing, I have preferred to use the characteristic $$ d $$ to denote ordinary differentials and the characteristic $$ \partial $$ for partial differentials. Adopting this convention rules out misunderstandings. So if $f$ is a function of $x$ and $y$, I will write
$$ df =\frac{\partial f}{\partial x}dx +\frac{\partial f}{\partial y}dy. $$

Whenever a function contains only a single variable, one may use the characteristic $d$ or $\partial$ indifferently. [...]

In order for the partial differentials, of a function which depends on more than one variable, to be completely determined, it does not suffice to provide the function to be differentiated and the variable with respect to which to differentiate; one must moreover express which quantities remain constant during the differentiation. For suppose $f$ is a function of $x,x_{1},\dotsc, x_{n}$. Take $n$ arbitrary functions $\omega_{1}, \omega_{2}, \dotsc, \omega_{n}$ of these variables and consider $f$ as a function of the variables $x,\omega_{1}, \omega_{2}, \dotsc, \omega_{n}$. Then, when $x_{1}, \dotsc, x_{n}$ remain constant, the $\omega_{1}, \omega_{2}, \dotsc, \omega_{n}$ will no longer be constant with changing $x$, and neither will, when $\omega_{1}, \omega_{2}, \dotsc, \omega_{n}$ remain constant, the $x_{1}, \dotsc, x_{n}$ stay constant. The expression $\frac{\partial f}{\partial x}$ will hence assume completely different values, depending on whether these or those quantities are assumed constant during differentiation.

(In modern terminology one could rephrase this as follows: the vector field $\frac{\partial}{\partial x}$ associated to a coordinate chart $x,x_1,\ldots,x_n$ on a manifold, does not depend on the coordinate function $x$ alone, but also on the remaining coordinate functions $x_1, \dotsc, x_n$.)

Suppose for example we introduce for a function $f$ of the two variables $x$ and $y$, an arbitrary function $u$ of $x$ and $y$, as the second independent variable instead of $y$. Then the differential that was previously denoted by $$ \frac{\partial f}{\partial x} $$ will now be expressed as $$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u}\cdot\frac{ \partial{u}}{{\partial x}}, $$ so that the same signs $\frac{\partial f}{\partial x}$ denote completely different values, depending on whether $y$ or $u$ are kept constant, while differentiating $f$ with respect to $x$.

I will therefore in this treatise, whenever partial differentials are needed, not only express with the statement: $f$ is a function of $x, x_{1}, \dotsc, x_{n}$, that $f$ depends on these variables, hence that it remains constant when these are constant, and changes, when they change — that would equally be valid, if instead of $x, x_{1}, \dotsc, x_{n}$ any other variables $\omega, \omega_{1}, \dotsc, \omega_{n}$ functions of these, were introduced as independent variables — rather when I say, $f$ is a function of the $x, x_{1} , \dotsc, x_{n}$ I want the following to be understood: whenever this function is partially differentiated, the differentiation should occur in such a way, that of these variables only one changes while the others remain constant.

Further, if the formulas are to be free of ambiguities, the notation should not only indicate the variable with respect to which the differentiation is occurring, but also the whole system of independent variables, the function of which is being differentiated, so that one may recognise which quantities remain constant during differentiation. And this is all the more necessary, since it is unavoidable that in the same calculation or even in one and the same formula, there appear partial differentials that refer to different systems of independent variables, for example in the above expression $$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u}\cdot \frac{\partial{u}}{{\partial x}} $$ in which $f$ is seen as a function of $x$ and $u$, while $u$ is seen as a function of $x$ and $y$; this expression was precisely the one $\frac{\partial f}{\partial x}$ transitions into, when $u$ is introduced as independent variable instead of $y$. But if we write next to the dependent variable all the independent variables to which the partial differentiation refers, then this expression can be depicted by the following formula, which is free of any ambiguity: $$ \frac{\partial f(x,y)}{\partial x} = \frac{\partial f(x,u)}{\partial x} + \frac{\partial f(x,u)}{\partial u}\cdot \frac{\partial{u(x,y)}}{{\partial x}}. $$

(To stress: he is using the same notation as function application, but he doesn't want to denote function application with it, he only wants to make explicit which variables are to be kept constant.)

Certainly this notation, as well as every other imaginable notation that would allow one to completely determine any partial differentiation from the symbols alone, would become very cumbersome in more general investigations or more involved formulas, yes even impracticable, since with higher numbers of independent variables and more terms it might happen that a formula, which can be expressed in a single line, takes up a whole page. Certainly one should place the highest value on a notation that eliminates any ambiguity, and which makes any formula understandable on its own, without oral clarifications. But when it was possible without too much disadvantage, and in view of the enormous and unavoidable verbosity of the notation, I settled for the shorter notation of differentials, that dispenses with the specification of the independent variables.

Up to this point one might argue that his notation is justified. But he starts doing it in places where I see no reason for it. For example further down he writes

Let $f, f_1, f_2$ etc. be mutually independent functions of the variables $x, x_1, \ldots, x_n$, and let $x$ be a variable contained in $f$ [...] it follows that the $m+1$ equations \begin{align} f(x,x_1,\ldots,x_n)&=\omega\\ f_1(x,x_1,\ldots,x_n)&=\omega_1\\ &\ldots\\ f_m(x,x_1,\ldots,x_n)&=\omega_m \end{align} do not determine more than $m+1$ quantities.

Here these $f$ are again variables quantities of type $\mathbb{R}$ and not of type $\mathbb{R}\to \mathbb{R}$, and there is no need to write $(x, x_1, \dotsc, x_n)$ next to the $f$'s in the equations.

I haven't been able to find similar "typing errors" in Bernoulli, Euler, Lagrange, Laplace, Gauss or Cauchy. Even after Jacobi there are several people like Riemann or Peano where I can't find this. Although it is easier to find in the second half of the 19th century. For example in Hermite or Maxwell and later in Felix Kleins lectures on mechanics as well as in Sommerfeld.

But since I have only looked at 1 or 2 works of each of the above authors, this answer is by no means conclusive.

  • $\begingroup$ If this answer dates the notation to Jacobi in 1840, what to make of the quote from Cajori's A History of Mathematical Notations, which points out that Euler in 1734 used it? He wrote both “si $f\left(\dfrac{x}{a} + c\right)$ denotet functionem quamcunque ipsius $\dfrac{x}{a} + c$.” (“$f\left(\dfrac{x}{a} + c\right)$ denotes some function of $\dfrac{x}{a} + c$”) and “Est vero $f(\dfrac{x}{a} + c)$ functio quaecunque ipsarum $a$ et $x$ nullius demensionis” — (contd) $\endgroup$
    – shreevatsa
    Commented Aug 17, 2017 at 17:14
  • $\begingroup$ — (contd) (“Now $f(\dfrac{x}{a} + c)$” is some function of $a$ and $x$ of zero dimensions”). So the very first time (according to Cajori) that parentheses were used to denote a function argument, the notation with the parentheses was also used as referring the function itself (the so-called “abuse of notation” that the question is about). Why does this not count? (Quoting from E045, original here, translation here) $\endgroup$
    – shreevatsa
    Commented Aug 17, 2017 at 17:14
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    $\begingroup$ @shreevatsa Eulers bracket notation (what we nowadays call function application) is not the abuse of notation. His $f$ in the quote is of type $\mathbb{R}\to\mathbb{R}$ so it's ok to apply it to something of type $\mathbb{R}$ and write say $f(1)$. But Jacobis $f$ is not of type $\mathbb{R}\to\mathbb{R}$ but of type $\mathbb{R}$ and it makes no sense to write $f(1)$ or the like. It might not be immediately obvious to you that Jacobis $f$ is of type $\mathbb{R}$ since he calls it a function. That's why I added the warning above, but I understand that I might need to provide more evidence. $\endgroup$ Commented Aug 17, 2017 at 19:55
  • $\begingroup$ @MichaelBächtold Yes his $f$ is of type $\mathbb{R} \to \mathbb{R}$, which means that for any $t$ like his $\dfrac{x}{a}+c$, his $f(t)$ is of type $\mathbb{R}$. My point is that he refers to $f(t)$ as a function, instead of referring to $f$ as a function. Is this not the same sort of abuse-of-notation you were asking about? (I see many people complain about this, and possibly jumped to the assumption that you were asking the same.) But on rereading, perhaps your question is solely about cases where (say) $y$ is of type $\mathbb{R}$, and someone writes $y(x)$ to denote that it depends on $x$. $\endgroup$
    – shreevatsa
    Commented Aug 17, 2017 at 20:01
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    $\begingroup$ @shreevatsa Euler is right in referring to $f(t)$ as the function, since that is the way the terminology was used before about 1900 (see the linked question in my answer and have a look at how Euler defined "function" in for example Institutiones Calculi Differentialis). I would agree with your last sentence, but would love to find a historical source who explains that they'll start writing $y(x)$ to denote that $y$ depends on $x$. Have you seen one? Certainly Jacobi doesn't explain it that way in De determinantibus. $\endgroup$ Commented Aug 17, 2017 at 20:37

I have seen this type of notation as a help in understanding quantifiers.

Example. Here is a statement (Bertrand's Postulate):

for every $k > 1$ there is a prime $p$ such that $k \le p < 2k$.

This may be written, to emphasize that $p$ depends on $k$, as:

for every $k > 1$ there is a prime $p = p(k)$ such that $k \le p < 2k$.

A reader can tell that $p$ depends on $k$ in the first one as it is. But putting $p = p(k)$ in the second one emphasizes that fact.

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    $\begingroup$ And we inadvertently applied the Axiom of Choice. $\endgroup$ Commented Aug 17, 2017 at 7:39
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    $\begingroup$ @AndrejBauer Where in this would AC be applied? For each $k$ we would just need to choose an element from a well-ordered set. $\endgroup$ Commented Aug 17, 2017 at 7:41
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    $\begingroup$ In this particular case you can avoid AC, but people do this sort of thing all the time, and in general it amounts to AC. $\endgroup$ Commented Aug 17, 2017 at 7:43

I don't feel that $\ f=f(x)\ $ is an abuse of notation. It is rather a message. When we have an expression like $\ f\ :=\ t^2\!\cdot\! x + s,\ $ then $\ f=f(x)\ $ means that in the future when we write $\ f'\ $ then it means $t^2$ and not $\ 2\cdot t\cdot x\ $ nor simply $\ 1.$ Otherwise, the announcement $\ f=f(x)\ $ doesn't really enter the rest of the proceeding. (Am I right?)

This notation and the calculations which follow it feels to us old-fashioned in the so-called pure mathematics because this kind of mathematical analysis appears much less these days than in the past.

We can still talk about an abuse, why not, but then there are many other convenient abuses.

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    $\begingroup$ Have you ever seen a historical (or modern) calculus textbook explicitly state this interpretation of $y=y(x)$? I might believe that this is how many people think of it, but this interpretation definitely abuses the standard mathematical meaning of $=$ and of function application. Also, thanks to Leibniz derivative notation it seems unnecessary. $\endgroup$ Commented Nov 1, 2016 at 18:37
  • $\begingroup$ It's not just about unnecessary but also about convenient. As we all know, notation $\ f'\ $ is convenient, it's simple. People liked it for centuries. (There is still another tricky moment about the notation involving functions and derivatives). $\endgroup$ Commented Nov 2, 2016 at 5:19
  • $\begingroup$ Responding to your last comment: there is definitely a place for $f'$-notation, namely when $f$ is a function in the modern sense. In that case it would be misleading to write $df/dx$ to denote $f'$. See this question: mathoverflow.net/q/115416/745 $\endgroup$ Commented Oct 25, 2023 at 7:12

Warning. This is an attempt at an answer out of curiosity rather than an expert answer.

Newton has the following passage in "Recomputation of surfaces of least resistance," (1694) (see Whiteside*, pp. 470-471):

Unde $aabb - 2aabx+aaxx+ bbxx = aay + xxy$

[capiendo fluxiones]

$- 2aab\dot x + 2aax\dot x+2bbx\dot x =2x\dot xy + aa\dot y+xx\dot y$

Whiteside (ibid) writes: "The dotted letters in immediate sequel are Newtonian fluxions; that is, $\dot x = \frac {dx}{dt}$ and $\dot y = \frac {dy}{dt}$ where t is some independent variable of ‘time’."

I'd like to add that I don't think that interpreted in the context (whether historical or modern), something like $x=x(t)$ (say in parametric equations) or $y=y(x)$ (say when** $y$ represents the distance from the $x$-axis at a certain $x$), would be an "abuse of the notation".

*The mathematical papers of Isaac Newton Volume VI 1684-1691

** I am pretty sure I have seen something like this in historical texts, but I couldn't remember where.

  • $\begingroup$ Thanks for you attempt to answer this, unfortunately I don't see why that quote of Newton is an example of what I'm asking for. Concerning you last paragraph, I understand that one may argue that writing y=y(x) is not technically an abuse of notation, but if you think of the y on the right as a distance, (which is not a function of type $\mathbb{R}\to\mathbb{R}$) then you are abusing the notation for function application. $\endgroup$ Commented Nov 1, 2016 at 15:56
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    $\begingroup$ I cannot see how Newton could calculate $\dot x$ without thinking of $x$ as $x(t)$, though he never wrote something like $x=x(t)$. In fact, that is why I added that comment, because the separation you have made would be only possible after having the modern definition of function. $\endgroup$ Commented Nov 1, 2016 at 16:19
  • $\begingroup$ Maybe this answer helps to understand the distinction between a function and a variable (which I find very relevant for such discussions): math.stackexchange.com/a/1259113/1984 $\endgroup$ Commented Nov 1, 2016 at 18:13
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    $\begingroup$ Newton is not so "explicit" but, due to his "dynamical intended interpretation" of the calculus, it is correct to say that the "independent variable" is time. See : Treatise of the Quadrature (1st ed 1710), page 1 : "considering that Quantities, which increase in equal Times, and by increasing are generated, become greater or less according to the greater or less Velocity with which they increase and are generated; 1/3 $\endgroup$ Commented Nov 2, 2016 at 13:08
  • $\begingroup$ ...I sought a Method of determining Quantities from the Velocities of the Motions or Increments, with which they are generated ; and calling these Velocities of the Motions or Increments Fluxions [$\dot x$], and the generated Quantities Fluents [$x$]." 2/3 $\endgroup$ Commented Nov 2, 2016 at 13:10

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