# Proving that special functions do not have closed-form expression

When dealing with special functions, like Erf, one should encounter the following statement

This function cannot be expressed in terms of classical functions

This seems pretty true, but I was wondering why. Could you prove it by hand ? I found the following definition on Wikipedia :

In mathematics, an expression is said to be a closed-form expression if it can be expressed analytically in terms of a finite number of certain "well-known" functions. Typically, these well-known functions are defined to be elementary functions—constants, one variable x, elementary operations of arithmetic (+ − × ÷), nth roots, exponent and logarithm (which thus also include trigonometric functions and inverse trigonometric functions).

I am not just interested in the error function but more generally in any special function. I was wondering if any "field" of mathematics related to this question existed. Any interesting reference of link would be appreciated !

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For $\operatorname{erf}$, see Liouville's theorem. – Argon Jul 26 '12 at 14:54
(grumble) Special functions are closed forms, it's just that their closed forms aren't elementary... – J. M. Jul 26 '12 at 15:46
@J.M. : well it depends on your definition of "closed-form expression". Here I am looking at, if you will, linear combination of elementary functions. In my applied field, a closed-form expression is a formula than can be computed using deterministic numerical methods, so yes, it depends on your point of view ;) – vanna Jul 26 '12 at 15:53
"a closed-form expression is a formula than can be computed using deterministic numerical methods" - by that token, then yes, the special functions are closed forms, except I'm not sure why you need the qualifying adjective "deterministic". Anyway, I've written at length on my feelings regarding the term "closed form" here. – J. M. Jul 26 '12 at 15:58
As you mentionned, the expression "closed-form" is context-dependent. I think I made my question clear to that regard : I would like to show that e.g. $erf$ cannot be expressed as a combination of constants, powers, $\exp$, $\log$, and so on. By the way one could say $erf$ is elementary ;) (this is pure trolling). – vanna Jul 26 '12 at 16:03

To be a bit more precise, the part of mathematics concerned with closed-form integrals and closed form solutions to ODE is called "Differential Algebra." This is not for the faint-hearted, and was begun by RITT and his student KOLCHIN at Columbia. One of the current leaders in this study is called "The Bozh" by my friend Dmitry, anyway Michael Boshernitzan of Rice. I have a book by Kaplansky called, and I think this is clever, "Differential Algebra."

A differential field is simply one which has a derivation, $D(uv) = u D(v) + v D(u).$ One can then decide whether a new item, an indefinite integral especially, is in the field. Thus the meaning of the phrase "closed form" is entirely up to the individual mathematician: a function is elementary if it is in the differential field being considered. Put another way, a function is elementary if you say it is. In the same way, if a child brings home a stray puppy, the parents tell the child not to give it a name as it will be gone soon. Your elementary functions are just those to which you have assigned a name.

There is a side note here: modern computer algebra systems use Ritt's algorithm, and many types of improvements based on including Grobner bases into the mix, for deciding whether some integral has a closed form, as far as the CAS is concerned. Note that one may do this in steps, first decide whether it is in the traditional basket of elementary functions we learn in high-school, next add in a few less traditional, see if that changes things.

There is a type of needlework called CROSS STITCH, for which a sampler is a piece small enough to be framed and hung on a wall. A book with text examples at BOOK. We could commission such a thing and send it to J.M. It would read

$$\begin{array}{c}\text{Elementary is in the}\cr\text{Eye of the Beholder}\end{array}$$

or maybe

$$\begin{array}{c}\text{Closed Form is in the}\cr\text{Eye of the Beholder}\end{array}$$

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Thanks for the historical background. And as for the cross-stitches, I should probably send J.M. one to apologize ;) – vanna Jul 26 '12 at 20:56
+1, of course. :D – J. M. Jul 26 '12 at 23:52

The "field" of mathematics that deals with such a question is, coincidentally, field theory. A function has a closed-form representation if and only if it belongs to a certain tower of function fields.

In practice, given a function $f: \mathbb{F}\to\mathbb{F}$, we want to determine if there exists an $n\in\mathbb{N}$ and a sequence of fields $\{K_i\}_{i=0}^{n}$ such that:

• $K_{i+1}$ is a simple field extension $K_i$ (described below)
• $K_0\equiv \mathbb{F[x]}$
• $f\in K_n$

The field extension must be of the form $K_{i+1}=K_i[g]$ where either:

• $g$ is algebraic over $K_i$
• $g=\exp(h)$ for $h\in K_i$
• $g=\ln(h)$ for $h\in K_i$

An elementary introduction can be found here

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a direct link to Goetz paper. – Raymond Manzoni Jul 26 '12 at 15:17