What is the general reason for functions which can only be defined implicitly?

Is this because they are multivalued (in which case they aren't strictly functions at all)?

Is there a proof?

Clarification of question.

The second part has been answered with example of single value function which cannot be given explicitly. The third part was automatically answered because there can't be a proof that necessarily implicit functions are all multivalued by way of the example of one that wasn't.

I don't think that the first part has yet been addressed. Stating that the answer can't be expressed in "elementary functions" seems tantamount to saying that it a necessarily implicit function, unless I'm missing something about the definition of "elementary functions". Such answers seem to imply that the equation could be solved in terms of "non-elementary" functions.If this is correct than I need to find out about them and how they could be used to calculate the dependent variable solely in terms of the independent one (my notion of an explicit function). This would seem to violate the notion of a function which could only be defined implicitly. I am also not concerned with whether or not the solution is closed or open form (by which I mean finite number of terms or infinite).

  • 1
    $\begingroup$ Perhaps you could give some examples of relations that can only be defined implicitly. $\endgroup$
    – Will Jagy
    Commented Jun 4, 2011 at 21:33
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    $\begingroup$ Whether a function has an explicit or closed form depends on what class of functions one admits as 'explicit'. Without a precisely specification of such a class of functions, your question cannot be answered. $\endgroup$ Commented Jun 4, 2011 at 21:37
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    $\begingroup$ The issue can be that there is no simple or elaborate combination of standard functions in terms of which the solution can be written. An example would be $e^y=xy$. One cannot "solve" for $y$ in terms of $x$ using only the so-called elementary functions. $\endgroup$ Commented Jun 4, 2011 at 21:44

4 Answers 4


Not necessarily. Consider the graph $G$ in ${\mathbb R}^2$ of the points $(x,y)$ such that $$ y^5+16y-32x^3+32x=0. $$ This example comes from the nice book "The implicit function theorem" by Krantz and Parks.

Note that this is the graph of a function: Fix $x$, and let $F(y)=y^5+16y-32x^3+32x$. Then $F'(y)=5y^4+16>0$ so $F$ is strictly increasing. Since $F(y)\to-\infty$ as $y\to-\infty$ and $F(y)\to\infty$ as $y\to\infty$, the equation $F(y)=0$ has at least one solution (by the intermediate value theorem), but then it has precisely one. This means that the graph $G$ describes a function (in the strict, traditional sense, not a multivalued function).

However, there is no formula (in the traditional sense) giving us $y$ explicitly as a function of $x$.

This specific example shows one of the reasons why this may happen. Here, we have $y$ as a solution of a quintic equation, but in general there is no algebraic formula that gives us these solutions; this is a deep result of Abel and Galois.

There are similar results stating that certain equations do not have elementary solutions (for example, certain differential equations), so they may define a function but we may not have a "formula" for it. In fact, the concept of what a "function" is underwent a few revisions before reaching its current modern form, precisely as we realized the need for versions more general than "given by an explicit formula". A nice account of the evolution of the term is in the paper by Penelope Maddy, "Does $V$ equal $L$?", J. Symbolic Logic 58 (1993), no. 1, 15–41. jstor, doi: 10.2307/2275321

  • $\begingroup$ Nice references! $\endgroup$
    – amWhy
    Commented Jun 5, 2011 at 3:01
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    $\begingroup$ $y^5+16y-32x^3+32x=0$. One may say that this is multi-valued, but the other four values are complex. $\endgroup$
    – GEdgar
    Commented Jun 5, 2011 at 3:22
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    $\begingroup$ The Abel-Galois result rules out an algebraic formula, not a formula involving general elementary functions. $\endgroup$
    – Johan
    Commented Dec 28, 2011 at 13:49

The notion of an "implicitly defined function" (IDF for short) has nothing to do with the question whether some such function can be expressed "in finite terms". The general reason for the appearance of an IDF is that we are given a relation $F(x,y)=0$ between two variables $x$ and $y$ (these may be vector-valued), where a priori these two variables have "equal rights". E.g., the relation $F(\cdot,\cdot)$ may be a natural law that connects the temperature $T$ and the pressure $p$ of some amount of gas in a closed vessel.

In such a case the question arises whether, given the value of $x$, the relation $F(x,y)=0$ determines a unique value $y$. If this is (maybe in a restricted sense, see below) the case, this value $y$ is a function of the chosen $x$; one is allowed to write $y=f(x)$ and says that the function $f$ is "implicitly defined" by the relation $F(x,y)=0$. If $F$ is given as an explicit expression in the variables $x$ and $y$ and one is lucky one may algebraically solve the equation $F(x,y)=0$ for $y$ and get an explicit expression for $f$. Usually one cannot solve for $y$, but one nevertheless is interested, e.g., whether $f$ is continuous, or one would like to know the value of $f'(x)$.

The theorem on implicit functions gives an answer to such questions. The essential point is that you need not to be able to solve your equation explicitly for variable $x$ and are nevertheless able to compute $f'(x)$ in individual points $x$. The theorem is a local theorem. It says that any point $(x_0,y_0)$ that satisfies the equation is the center of a rectangular window $W=U\times V$ such that the set $\{(x,y)\in W | F(x,y)=0\}$ can be viewed as the graph of a "local" function $f:\ U\to V$. There is a formula for $f'(x_0)$, namely $$f'(x_0)=-{F_x(x_0,y_0) \over F_y(x_0,y_0)} \ .$$ This formula shows that we forgot to mention the main assumption for all this to hold: One needs $F_y(x_0,y_0)\ne 0$.


Explicit functions are great for expressing a dependent variable as a function of an independent variable. But it's not always possible, and even if possible, not always most appropriate to arbitrarily define one variable in terms of another (or in terms of many).

Some implicitly defined functions (Ex 1), but not all (Ex 2), are not (strictly speaking) functions at all.

  • Ex 1 (not a "strict" function): $f(x, y) = \{(x, y) | x^2 + y^2 = r^2, r>0\}$ where $r$ is some positive constant, defines a circle of radius $r$. It is not a function in the strict sense of the term, since, if $(x, y)$ is a solution, then so is $(x, -y)$. (Or alternatively, if $(x, y)$ is a solution, then so is $(-x, y)$). However, we can define each of the semicircles for $y \geq 0$ and for $y < 0$ as explicit functions, such that the graphs of both functions forma circle.

  • Ex 2 (strict function expressed implicitly): There are some strict functions which are best expressed as implicit functions, or for which one variable cannot be expressed explicitly as a function of another variable using only elementary functions, like $$e^y = \frac {xy}{x+y}$$

Rather than elaborate here, Wikipedia does a nice job of clarifying the distinction between implicit and explicit functions, and/or when one form is preferable to the other:

I also encourage you to take a look at the Implicit Function Theorem:

In multivariable calculus, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.

  • $\begingroup$ I have no difficulty with anything that you've said.-amWhy. I am looking for some characteristic of functions that can only be described implicitly that separates them from those that can be described implicitly. $\endgroup$ Commented Jun 6, 2011 at 5:39

Just to give an example that occurs in applications, consider the Kepler equation


relating the eccentric anomaly $E$ of a body orbiting on an elliptic path of eccentricity $\varepsilon$ (thus, $0 < \varepsilon <1$) as a function of the mean anomaly $M$. It can be seen here that it is not too straightforward to determine $E$ as an explicit function of $M$, even though $E$ is indeed a function of $M$. However, there is an infinite series solution:

$$E=M+2\sum_{k=1}^\infty \frac{\sin(k M)J_k(k\epsilon)}{k}$$


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