In high-school math functions always looked to me just like glorified equations. The only time I saw a meaningful difference was when we covered the equation of a circle and I realized that an equation can describe a circle but a function cannot (a function can describe only one half of the circle).

So what is the difference between a function and an equation that leads to this, in formal terms?

  • 1
    $\begingroup$ The implicit function theorem is relevant, but it is beyond high school math. $\endgroup$
    – Git Gud
    Jan 21, 2015 at 17:04
  • 12
    $\begingroup$ A function has to pass the "vertical line test." It can at most have one point at any vertical line. $\endgroup$
    – Eff
    Jan 21, 2015 at 17:05
  • 24
    $\begingroup$ Why can a circle be described by an equation, but not by a function? - But it can be described by a function! Just let $r(\theta)=$ constant, and plot it using polar coordinates instead of Cartesian ones. ;-$)$ $\endgroup$
    – Lucian
    Jan 21, 2015 at 20:03
  • $\begingroup$ @Eff That's only true for single-valued functions. You can allow less strict definitions for appropriate situations (but be aware of which theorems then fail, of course). $\endgroup$ Jan 22, 2015 at 13:25
  • $\begingroup$ @Lucian: Of course, in polar coordinates the "vertical line test" becomes a "radial line test". Which is why y=exp(x) can't be expressed as a function r=f(θ) $\endgroup$
    – MSalters
    Jan 23, 2015 at 10:44

5 Answers 5


Well, a circle can be described by a function, just not in the sense that you may be familiar with. If you are looking at a function that describes a set of points in Cartesian space by mapping each $x$-coordinate to a $y$-coordinate, then a circle cannot be described by a function because it fails what is known in High School as the vertical line test.

A function, by definition, has a unique output for every input. However, for almost all points on a circle, there is another point with the same $x$-coordinate. So, you would need your function to give two different $y$-coordinates for certain inputs, which is not allowed.

However, there is no rule that the input of a function has to be an $x$-coordinate or that the output has to be a $y$-coordinate, so we can define other functions that describle a circle. In more formal terms, the domain and codomain of a function do not have to be $\Bbb{R}$. For example, we can have a function that outputs an ordered pair (that is, codomain of $\Bbb{R}\times\Bbb{R}$). Then, $$f(t)=(\sin t,\cos t)$$ outputs the unit circle when $0\le t<2\pi$. We could also describe the points in space in a different way, using polar coordinates. Here we use the counter-clockwise angle from the positive $x$-axis, $\theta$, and the distance from the origin, $r$, to identify a point. Using this system, we can easily describe the unit circle as $(\theta,f(\theta))$, where $f(\theta)=1$ and $0\le\theta<2\pi$.

  • 6
    $\begingroup$ This answer makes me realize that the question isn't so simple as it seems. $\endgroup$
    – Siyuan Ren
    Jan 22, 2015 at 1:32
  • $\begingroup$ @SiyuanRen And this is just assuming a naive interpretation of a function "describing" a circle, without getting into what that could mean. Note that the points of a circle are the image (output) of the first function, but they are the ordered pairs of input and output of the second. $\endgroup$
    – KSmarts
    Jan 23, 2015 at 21:19
  • $\begingroup$ I think the word unique should be replaced by single. $\endgroup$
    – Shub
    Jun 20, 2021 at 15:06

You say that "functions look like glorified equations". There's definitely some truth to that. Here's an equation:

$$y = x^2$$

Here's another equation:

$$f(x) = x^2$$

Both of these are equations, and the two equations do very similar things. The two equations both define functions. But they do it a little bit differently.

The first equation doesn't give a name to the function it defines; it just defines $y$ as being "a function of" $x$, which is to say that it tells you what $y$ is once you know what $x$ is. The second equation does give a name to the function it defines; the function is called $f$. (The function isn't $f(x)$; the function is just $f$.)

Now, here's another equation:

$$x^2 + y^2 = 1$$

Unlike $y = x^2$, this equation does not define a function. Why not? Because it doesn't tell you what $y$ is once you know what $x$ is. This equation defines a relation, which is to say that it tells you what the values of $x$ and $y$ are allowed to be, but the value of $y$ is not completely determined by the value of $x$.

A function, it turns out, is just a special kind of relation. A function is any relation that has the property that once you know what the first value is, you know what the second value is. A circle can be described by a relation (which is what we just did: $x^2 + y^2 = 1$ is an equation which describes a relation which in turn describes a circle), but this relation is not a function, because the $y$ value is not completely determined by the $x$ value.

Now, could we use something similar to function notation in order to define a circle? Sure. What we can't do is something like this:

$$x^2 + f(x)^2 = 1$$

Since we're using function notation here, it looks like we're still trying to define a function. But what we can do is give our relation a name. Let's call it $\diamond$. Now we can say this:

$$x \diamond y \text{ whenever } x^2 + y^2 = 1$$

Now, much like $f$ is the name of a function defining a parabola way above, $\diamond$ is the name of a relation defining a circle.

  • 6
    $\begingroup$ Now that you say "x⋄y whenever x2+y2=1", it occurred to me that you could define any equation as a function taking several parameters, which then returns true/false for every combination of parameters. $\endgroup$
    – sashoalm
    Jan 22, 2015 at 8:36
  • $\begingroup$ On the other hand, we can define a function by $\;\;\; f(x)+(\hspace{.04 in}f(x))^{\hspace{.02 in}5} \: = \: x \;\;\;\;$. $\;\;\;\;\;\;\;\;\;$ $\endgroup$
    – user57159
    Jan 22, 2015 at 13:39
  • $\begingroup$ Well, i would argue that the second equation is still not a function because you are missing the domain and the codomain. $\endgroup$ Jan 22, 2015 at 15:51
  • 2
    $\begingroup$ @sashoalm Yes, and that function is called the indicator function for the set of pairs $(x,y)$ that satisfy the relation. $\endgroup$
    – becko
    Jan 22, 2015 at 16:55

This is sort of an artifact of the way we draw functions. We place (input, output) pairs on the plane through the arbitrary process

  1. Start at the center of the paper.
  2. Travel [input] distance to the right
  3. Travel [output] distance up
  4. Place a dot on the paper.
  5. Repeat for all possible inputs

You can see that this process can never draw a circle because at step 3, we have to go either up or down depending on whether [output] is positive or negative. But for a circle we would have to do both!

Imagine instead we did this instead:

  1. Imagine a horizontal line through the center of the paper
  2. Imagine a ray with vertex in the center of the paper, making an angle of [input] from the horizontal line
  3. Travel [output] distance along the ray
  4. Place a dot on the paper
  5. Repeat for all possible inputs

Then a circle of radius $r$ would very easily be represented by the function $f(x) = r$. (This is commonly known as polar coordinates)

Lastly, the way we usually draw equations is this:

  1. Pick a point on the paper
  2. If the point's horizontal and vertical distance to the center of the paper satisfy the equation, draw a dot on the paper.
  3. Repeat for all points on the paper

You can see that this method does not have the same restriction as the first method, and therefore the image of a circle could result.


When the relation is univocal (one $x$ gives one $y$), you can use a function, like $y=f(x)$.

When the relation is not univocal (one $x$ gives several $y$), you need several functions, like $y_0=f_0(x),y_1=f_1(x)$.

This can be combined in a single equation like $(y-f_0(x))(y-f_1(x))=0$.

In the case of a unit circle, for example, $$y_0=\sqrt{1-x^2},\\y_1=-\sqrt{1-x^2},$$ or $$(y-\sqrt{1-x^2})(y+\sqrt{1-x^2})=0,$$ i.e. $$x^2+y^2=1.$$

More generally, an implicit equation like $F(x,y)=0$ gives several $y$ for one $x$ and several $x$ for one $y$.


If you allow functions to misbehave a bit (multi-valued functions) like square roots for example, then a circle can be described as a function. (e.g. the square root of 9 is both +3 and -3 (since both values, when squared, yield 9)).

E.g. solving $x^2 + y^2 = r^2$ for y:

$y = \sqrt{r^2 - x^2}$

Outside the circle (i.e. when x < -r or x > r), y is undefined, but within the circle, the square root above has two roots forming the top and bottom parts of the circle.

Hmmm - I don't have enough rep to comment, hence the answer.

  • 2
    $\begingroup$ The standard definition of the square root function is that it evaluates to the positive square root. $\endgroup$
    – user856
    Jan 22, 2015 at 22:09
  • 1
    $\begingroup$ Sure, although both positive and negative roots are in fact square roots, hence the need to "allow function to misbehave a bit" :) See merriam-webster.com/dictionary/square%20root and mathworld.wolfram.com/SquareRoot.html $\endgroup$ Jan 26, 2015 at 18:55
  • $\begingroup$ @Rahul "The standard definition of the square root function is that it evaluates to the positive square root.". Actually that's not true. Wolfram MathWorld says "Note that any positive real number has two square roots, one positive and one negative". Merriam-Webster says "the square root of 9 is ±3. Wikipedia says "For example, 4 and −4 are square roots of 16 because 4^2 = (−4)^2 = 16." $\endgroup$ May 17, 2017 at 19:51
  • $\begingroup$ Any positive real number has two square roots, but as I said, the square root function evaluates to the positive square root, otherwise it would not be a function. $\endgroup$
    – user856
    May 17, 2017 at 20:57
  • $\begingroup$ Ok yes I didn't notice. You are right, thank you. $\endgroup$ May 18, 2017 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.