What is really the TRUE definition of an implicit function?

Question

First of all I would like to say that I have already found similar questions on stack exchange but somehow my confusion regarding the definition of an implicit function still linger.

The title says it all, but here's the question: what is really the TRUE definition of an implicit function?

I've scoured the internet and books for definitions and what I've found is that the word implicit function is used/defined in two ways:

1. An implicit function is an equation involving two variables (e.g., x and y) that is possible to solve for y in terms of x but is sometimes hard/messy/impractical. An example of an implicit function using this definition is $x^2+y^2=1$. (some sources: MITx, statisticshowto)

2. Given a relation of the form $f(x,y,)=g(x,y)$ where $f$ and $g$ are functions, an implicit function is any function we get by taking the relation f(x,y,)=g(x,y) and solving for y. An example of an implicit function using this definition would be: $y=+\sqrt{(1-x^2)}$ and $y=-\sqrt{(1-x^2)}$ from the relation $x^2+y^2=1$. The functions $y=+\sqrt{(1-x^2)}$ and $y=-\sqrt{(1-x^2)}$ are also called the implicit function of the relation $x^2+y^2=1$. Some implicit functions from a relation can also be impractical/hard/messy to find. (Source:dartmouthpdf, definition clipped from dartmouth pdf).

This is my confusion: on one hand, definition 1. says that an implicit function is the equation $x^2+y^2=1$. On the other hand, definition 2. says that an implicit function is the function that we get by isolating y from a relation of the form $f(x,y)=g(x,y)$ where $f$ and $g$ are functions (e.g., $y=+\sqrt{(1-x^2)}$ and $y=-\sqrt{(1-x^2)}$ from the relation $x^2+y^2=1$)

On one hand, some resources refer to the $x^2+y^2=1$ as the implicit function. On the other, some refer to $x^2+y^2=1$ as a relation which has two implicit functions: $y=+\sqrt{(1-x^2)}$ and $y=-\sqrt{(1-x^2)}$.

Thank you for reading my question.

Answer 1

The "true" definition of an "implicit function" is that it's shorthand for the idea of an implicit definition of a function.

All of your sources approach the right idea but express it in poor language. "Implicit" and "explicit" are properties of the definition of the function, not the function itself. The more precise language is to say that a function is implicitly defined. Calling something an implicit function is really shorthand for saying that the function is implicitly defined. Calling an equation an implicit function is particularly sloppy nomenclature: an equation can be used to define a function, but it is not itself a function, implicit or otherwise! The correct nomenclature is that an equation can provide a definition of a function, and that definition is either implicit or explicit.

Let's stick to 1 variable for simplicity. A function $f$ of one variable is implicitly defined if it is defined through a relationship between the values $f(x)$ and $x$, but that relationship need not directly give an equality between $f(x)$ and some other already known function of $x$. A common way where $f$ is implicitly defined is if $f(x)$ is defined through another function $g$ of two variables, e.g. $g(x,f(x)) = 0$. As an example, let's take $g(x,y) = x^2 + y^3$. Then the equation $g(x,f(x)) = 0$ reads $$ x^2 + (f(x))^3 = 0. $$ This equation doesn't immediately tell you how to obtain the values of $f(x)$. It instead relates the values of a function of $f(x)$ to another function of $x$. Now, in this case I have chosen $g$ specifically so that you can solve for $f(x)$: $$ f(x) = -x^{2/3}. $$ This expression provides an explicit definition of $f$ as a function of $x$. If I have $x$, I can plug it directly into this expression to get $f(x)$ without doing any further manipulation: that's what makes it explicit. The earlier expression $x^2 + (f(x))^3 = 0$ does not constitute an explicit definition of $f(x)$ because instead of telling you directly what $f(x)$ is, it tells you indirectly by describing how $f(x)$ relates to other functions of $x$. Therefore we say that it constitutes an implicit definition instead.

Of course, the expression $f(x) = -x^{2/3}$ is also abstractly just a relationship between $f(x)$ and $x$ (albeit a very straightforward one), so you could also say that this expression constitutes an implicit definition of $f$ as a function of $x$. It just happens to also constitute an explicit definition as well. An explicit definition of $f$ is also trivially an implicit one, but not all implicit definitions are explicit.

So in the context of your examples, here's the more precise terminology. An expression of the form $F(x,y) = 0$ (which includes $f(x,y) = g(x,y)$) gives an implicit definition of $x$ as a function of $y$, or $y$ as a function of $x$, assuming that you can indeed solve those equations for unique solutions. The earlier example with $g(x,y) = x^2 + y^3$ is a case in point. Declaring that $(x,y)$ satisfies $$ x^2 + y^3 = 0 $$ constitutes an implicit definition of $y$ as a function of $x$, which you can solve to obtain the explicit definition $y(x) = -x^{2/3}$. It also constitutes an implicit definition of $x$ as a function of $y$. However, in this case we run into two technicalities:

1. Due to the square root, if we try to write $x$ as an explicit function of $y$, we actually end up with two functions: $x(y) = \pm\sqrt{y^3}$. That is, the implicit definition $x^2 + y^3 = 0$ fails to uniquely define $x$ as a function of $y$.
2. Regardless of which version of $x(y)$ we choose, we need to restrict the domain to $y\geq 0$ (assuming we wish to avoid complex numbers).

As we can see here, there can be a few exceptional scenarios when converting an implicit definition of a function to an explicit one. (Edit: the implicit function theorem mentioned in the other answers is basically formulated to let us identify and address these exceptional scenarios.)

In practice, we use the terms "explicit function" and "implicit function" with the above understanding, as this is shorter to write and say.

Answer 2

One version of the implicit function theorem is the following:

Theorem. If $f(x,y)$ is a smooth function defined on an open set $U$ in the plane, and $\frac{\partial f}{\partial y}(x_0,y_0) \ne 0$ at a point where $f(x_0,y_0) = 0$, then for a small neighborhood of $x$ near $x_0$, we can find a (unique) smooth function $y = g(x)$ such that $$f(x,g(x)) = 0$$ holds.

It's reasonable to consider the function $y = g(x)$ as a function "implicitly" defined by the relation $f(x,y) = 0$.

Example: Take $f(x,y) = x^2 + y^2 - 9$. At the point $(0,-3)$, $f(0,-3) = 0$ and $$ \frac{\partial f}{\partial y}(0,-3) = 2\cdot 0 + 2\cdot (-3) = -6\ne 0, $$ and as we know, we can express $y = g(x) = -\sqrt{9-x^2}$ for values of $x$ near $0$ in such a way that $$ f(x,g(x)) = 0. $$ The function $y=g(x)$ we obtain as guaranteed by the implicit function theorem is what I would consider as an implicit function in the classical calculus sense of the term.

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Added: To state things a little less precisely, but with more intuition, if $f(x,y) = 0$ is an equation involving $x$ and $y$ (often referred to as a relation), and if we can "solve the equation $f(x,y) = 0$" for $y$ as a function of $x$—meaning we can find a function $g(x)$ so that $$f(x,g(x)) = 0,$$ then we call $g(x)$ an implicit function defined by the relation $f(x,y) = 0$. In other words, we can regard the relation $\{(x,y):f(x,y)=0\}$ as the graph of the function $g(x)$.

The implicit function theorem I stated in the first part of my answer gives a precise condition we can check (based on the partial derivative $\partial f/\partial y$) to tell whether we can actually solve for $y$ as a function of $x$ in the relation $f(x,y) = 0$.

Sometimes there is more than one implicit function defined by the relation $f(x,y) = 0$—any function $g(x)$ or $h(y)$ that satisfies $f(x,g(x)) = 0$ or $f(h(y),y) = 0$ can be regarded as an implicit function defined by the relation $f(x,y) = 0$. This is a little subtle, but the key is that they are all locally defined functions. There may be no single function $g(x)$ whose "$y$-values" cover the whole relation $f(x,y) = 0$. For example, in the circle equation $x^2+y^2-9 = 0$, we need four (at least two) implicit functions to cover the entire relation, $x = \pm\sqrt{9-y^2}$ and $y=\pm\sqrt{9-x^2}$. All four of these equations define implicit functions satisfying the circle relation, though we can artificially get more implicit functions by restricting the domains of these four functions.

You'll notice in the statement of the implicit function theorem, and the example I gave, we single out a unique implicit function by considering points $(x,y)$ near to some basepoint $(x_0,y_0)$ (in the example, $(x_0,y_0) = (0,-3)$, but if we started with $(0,3)$ instead, we would have found a different valid implicit function).

You will typically see a proof of the implicit function theorem and a discussion of its important role in mathematics in a Real Analysis class in college.

Answer 3

Mathematics does not have a central authority to decide on "official" definitions of terms, so different authors may sometimes use different definitions. But what I would say is that an implicit function is a function $y(x)$ that satisfies an equation $f(x, y) = g(x,y)$, i.e. satisfies $f(x,y(x)) = g(x,y(x))$ on its domain.