# What is the mathematical relevance of whether an expression has a closed form?

In the evaluation of mathematical expressions, particularly integrals, I often find a statement that the expression has or does not have a closed form. I looked up the definition, and the important point seems to be that closed forms use a finite number of elementary operations, where "elementary" includes exponentiation, logarithms, and trigonometric functions.

I understand that an infinite sum is somehow "not closed", e.g. in the sense that its actual numerical value can only be computed up to a certain precision; and I also understand that $\exp()$ is somehow elementary. However, the actual numerical value of $\exp(x)$ can also only be computed approximately because the function is defined by an infinite sum. So isn't this just a trick, declaring exponentiation elementary and therefore "closed" while it's actually just as infinite and therefore "open" as any other function defined by an infinite sum?

In particular I'm thinking of the error function, $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt$. How is $$\exp(x) = \sum_{n = 0}^{\infty} {x^n \over n!}$$ more elementary than $$\operatorname{erf}(x)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^n ~ x^{2n+1}}{n! ~(2n+1)}$$ ?

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The intuitive content is: "There is no neat way of writing down the solution in terms of functions we feel comfortable about, so don't bother looking for such a way." –  Michael Greinecker Feb 7 at 13:27
@MichaelGreinecker: Well I guess then the question for my particular example becomes: Why do we feel uncomfortable about erf($x$)? Not that I'd question this: I do feel uncomfortable about it. But maybe only because some professor told me that it doesn't have a closed form? ;-) –  A. Donda Feb 7 at 13:33
@A.Donda the other thing that is often implied by closed form is that there is an explicit solution available (even if it is in terms of elementary functions where series or lookup tables etc have to be used). If the dependent variable is only defined implicitly (especially if you can't even solve the differential equations) then you have to go and do some kind of search or iterative procedure to find the solution, with associated problems of knowing whether you have found the right solution, instability etc. –  TooTone Feb 7 at 15:00
@TooTone: In this sense erf($x$) is a closed form solution, right? Because the series counts as "explicit"? –  A. Donda Feb 7 at 15:16
@A.Donda yes I'd say that erf is a closed form solution. In fact there are some very well known formulas (e.g. Black-Scholes in finance) which involve erf which are closed form. I think there's a certain amount to "closed form" which it technical (as covered by the answers below), but also a certain amount which is a bit more about common parlance, which I tried to cover in my comment. It's a good question anyway! –  TooTone Feb 7 at 15:18

As others have pointed out this is mostly a matter of convention, but the convention is not completely arbitrary. Note that it has nothing to do with the numerical approximation or infinite series; by that standard you should be just as unhappy with $\sqrt{x}$ or $\sin(x)$ as you are with $e^x$. (Indeed, rational functions are the only functions whose values can be calculated without infinite series.)

Instead, the convention has to do with differential equations. The reason why $e^x$, $\sin x$, and $\cos x$ are considered elementary is that (together with polynomials) they are the only functions you need to write down solutions to differential equations of the form $Dy = 0$ where $D$ is a linear differential operator (i.e. a polynomial in $\frac{d}{dx}$). Since linearization is is the most basic tool in the theory of differential equations in general, exponential functions and trigonometric functions play a central role. Logarithms (and inverse trigonometric functions) appear as inverses of elementary functions.

This discussion is the starting point for a larger theory. One defines a differential field to be a field $F$ equipped with a derivation $d$; the standard example is the field of rational functions equipped with differentiation. Just as one extends ordinary fields by adjoining a root of a polynomial, one can extend differential fields by adjoining a solution to an equation involving the derivation. It turns out that two specific differential field extensions are most fundamental:

• Exponential: adjoin a solution to $dy = y \cdot df$ for $f \in F$
• Logarithmic: adjoin a solution to $dy = \frac{df}{f}$ for $f \in F$.

This language can help answer questions about the structure of solutions to linear differential equations (for instance, which functions have elementary antiderivatives).

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Thanks a lot for this answer! In my research I found some references to differential algebra, but none of the explanations were as clear as yours. To summarize: exp, log, sin & friends are "elementary" in the sense that they occur naturally in extending sets of functions by solutions to differential equations, which are expressed using functions already in the set. Then, is there another common extension of differential fields such that e.g. erf($x$) would become an element, considering it is the solution to a differential equation than can be written in terms of elementary functions? –  A. Donda Feb 7 at 15:27
I agree that my attempted explanation using numerical approximation is misguided. It was motivated by the "finite" in the definition of "closed form". But yes, on the same grounds I'd question $\sqrt{x}$ to be elementary. How does the "finite" part fit in with the differential algebra motivation? –  A. Donda Feb 7 at 15:34
Here's how $erf$ seems to fit in this framework. Begin with the differential field $(\mathbb{C}(x), \frac{d}{dx})$ of rational functions and form the exponential extension $F$ corresponding to $dy = y \cdot d(-x^2)$. Expressing the solution as $y = -2x e^{-x^2}$ and multiplying by the rational function $\frac{-1}{2x}$, we see that $f(x) = e^{-x^2}$ lives naturally in $F$. The function $erf(x)$ would naturally live in the extension of $F$ obtained by adjoining a solution to the equation $dy = f$. –  Paul Siegel Feb 7 at 21:39
However, the latter extension is neither exponential or logarithmic, and thus it doesn't have the nice properties enjoyed by those extensions. –  Paul Siegel Feb 7 at 21:43
@A.Donda RE your other question, one defines an elementary function to be any function which lies in a finite tower of exponential or logarithmic extensions of $(\mathbb{C}(x), \frac{d}{dx})$. For instance, if you adjoin a solution $y_1$ to the equation $dy_1 = \frac{dx}{x}$ (i.e. $y_1(x) = \log(x)$) and then adjoin a solution $y_2$ to the equation $dy_2 = y_2 \cdot d(\frac{1}{2}y_1)$ then you get an extension which contains $e^{1/2 \log x} = \sqrt{x}$. So $\sqrt{x}$ is elementary because it lies in an exponential extension of a logarithmic extension of $(\mathbb{C}(x), \frac{d}{dx})$. –  Paul Siegel Feb 7 at 21:53