# Is there a difference between $y(x)$ and $f(x)$

Oftentimes functions described by $f(x) = 2x+4$, and when this is mapped to the Cartesian plane, $f(x) = y$. This surely implies that $y = 2x+4$. Is there a difference between this and $y(x) = 2x+4$?

## 6 Answers

### Functions vs. coordinates

Consider plots of two different functions: $f(x)$ and $g(x)$ on the same $xy$ plane. One curve will be labeled $y=f(x)$ which means "this is a set of $(x,y)$ points that satisfy $y=f(x)$ condition". The other will be labeled $y=g(x)$.

Which of these should define $y(x)$? Both? – certainly not, because $f$ and $g$ are different functions. I say: neither. The statement $y=f(x)$ is just a condition for some set of points (i.e. $(x,y)$ pairs) while $y=g(x)$ is another condition for another set of points.

Explicit definition in a form $y(x)=…$ does define a function (well, does or doesn't, read the next paragraph). In this case $y$ is just an arbitrary name and may replace $f$. The same symbol $y$ may be a coordinate on $xy$ plane, which was $xf$ plane before the name replacement. (It is only a custom to have $xy$ plane.) This "union" of function name and coordinate name may cause a problem when there is another function $g(x)$ to plot.

It should be obvious that if $y$ replaces $f$ it cannot replace $g$ that is different than $f$.

For that reason it is a good thing to have coordinates with symbols which are not function names.

### Definitions vs. equations or conditions

Another problem: we often write function definitions the same way as conditions to be met or equations to solve. Compare the two:

$$cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ $$cos(x) = \frac{1}{2}$$

The former may be treated as non-geometric definition of $cos$ function. The latter is just the equation to solve for $x$. We have some experience and often feel the difference, but a person (say: Bob) completely unaware of $cos$ will be confused. Bob may find every $x$ that satisfies

$$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = \frac{1}{2}$$

and still will not be able to tell what number $cos(x)$ is equal to for any other $x$.

It's worse than that! Bob cannot tell what number $cos(x)$ is equal to even for $x$ being his solution, because he cannot be sure that either equation defines the function (we know it's the first one, Bob doesn't). To clarify that, let's see what happens when I change $cos$ to $sin$ only:

$$sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ $$sin(x) = \frac{1}{2}$$

We know by experience that neither of above defines $sin$. Yet these are legitimate equations to solve either separately or as a system (with empty set solution). Bob (not knowing about $sin$) may only assume that one of the equations is a definition – this will be wrong, his set of solutions will not be empty.

That's why I like the notation $f(x) \equiv …$ or the word "def" above the equality sign, or the explicit statement ("let us define…") – just to cut out possible ambiguity.

I've got the impression that you meant $y(x) \equiv 2x+4$ because there is no other expression in your example that you may want to define as $y(x)$.

### Summary

1. Is there a difference between $y=2x+4$ and $y(x)=2x+4$?

• My answer is: in general it may be. The second form is more likely to be read as a definition of a function, yet any form may or may not be intended to be a definition. Both may be equations to solve, when $y(x)$ is defined elsewhere ($y$ may be a given number or a parameter not depending on $x$, still it can be formally written as $y(x)$). The second form states that there is some function $y(x)$; the first one may mention a function or a variable (coordinate) $y$. The coordinate (not function) interpretation allows the first form to be a condition for points (i.e. $(x,y)$ pairs) – as it may be in your example – that leaves room for another conditions for another sets of points.
2. Is there a difference between $y=2x+4$ and $y(x) \equiv 2x+4$?

• Yes. The second form defines a function for sure. The first one may have another meaning (explained above).
3. Is there a difference between $y \equiv 2x+4$ and $y(x) \equiv 2x+4$?

• There is a subtle one: from the second form we know the independent variable is $x$; it may be $x$ xor $y$ in the first one.
4. Is there a difference between $y(x)$ and $f(x)$?

• No. In a sense: you can name your function with any unused symbol. But if $y$ is already in use (e.g. to name a different function, coordinate, parameter) then you cannot freely rename $f$ to $y$.
• This answer was very illuminating to me, especially since it was on a topic that I thought I was comfortable with. I particularly enjoyed your 2nd section, highlighting the important difference between equality and definitions/the distinctness of objects themselves. It seems like $:=$ "$\subset$" $=$. In other words, the definition symbol can serve as an equality sign. But the equality sign should not be used to stand for definitions/distinctness. I will probably start using the notation $f :=$, $f \equiv$, or $f: x \mapsto$ more frequently for defining functions. Thanks again – DWade64 Oct 3 '18 at 13:34

$y(x)\equiv f(x)$ if and only if $y$ is the name given to the dependent variable. In general, $f(x)$ could be named anything but $x$. Consider $z=f(x)$ which is a single variable function that is the $xz-$trace of a function $f(x, y)$. In this scenario, $y=f(x)$ is the trace in the $xy-$plane of the same two-variable function.

(Perhaps to be even more specific we'd have to call the traces something like $f$, $g$, and $h$ to avoid ambiguity.)

Basically they means the same thing, because $y$ is a variable which depend only by $x$.

• Only in certain context. What is there to say about $y(t)$ or $y(\theta)$? – John Molokach Jul 1 '16 at 0:32

Actually, they are the same thing, however $y(x)=2x+4$ is more complete than $y=2x+4$ as a written statement because when you see $y(x)$, then you know $y$ being a function of $x$ before considering the right side.

It is an interesting question. I think they are the same. You may write $y=\cdots$ or $y(x)=\cdots$.

Actually, I think everyone will say that they are the same, and in fact I am wondering if someone answer to this question that they are not!

No, there's no difference. But you don't really see $y(x)$ all that much since it is commonly indicative when you say $y$ or $f(x)$ and then have $x$ in your equation on the Cartesian plane.

Although, I suppose if you don't want to confuse people, you can say $y(x)$ to indicate that it's solely on the $xy$ Cartesian plane and not the $yz$ or $xz$ Cartesian planes. But that would be repetitive if you say $y=2x+4$ and you can clearly see that there are only two variables.

The $f$ in $f(x)$ is just a notation (most, if not all the time) unless it is very clear that it's the $f$-axis, which is incredibly uncommon.