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$\newcommand{\lax}{\operatorname{lax}}$ Liouville's theorem is well known and it asserts that:

The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions.

The problem I got from this is what is an elementary function? Who defines them? How do we define them?

Someone can, for example, say that there is a function which is called $\lax(\cdot)$ which is defined as:

$$ \lax\left(x\right)=\int_{0}^{x}\exp(-t^2)\mathrm{d}t. $$

Then, we can say that $\lax(\cdot)$ is a new elementary function much like $\exp(\cdot)$ and $\log(\cdot)$, $\cdots$.

I just do not get elementary functions and what the reasons are to define certain functions as elementary.

Maybe I should read some papers or books before posting this question. Should I? I just would like to get some help from you.

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We can and we do. Check out wolfram's function library. –  David H Jun 16 at 20:27
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The intention of the term is that certain functions are used as "elements" for building up more complicated functions by the use of arithmetic operations, function composition, and so forth. But the category can be viewed as just a convention: powers of variables, polynomials, exponential and logarithmic functions, trigonometric and inverse trigonometric functions are the ones we knew best from back when the term "function" began to be used consistently. We could always add others to the "list". –  RecklessReckoner Jun 16 at 20:34
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What is called $\operatorname{lax}(\cdot)$ here is more commonly known as the error function $\operatorname{erf}(\cdot)$ (up to a constant factor). –  Arthur Jun 16 at 20:40
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Would the Special Functions tag be appropriate here? :) –  Shaun Jun 17 at 19:05
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... of course, then $\int \mathrm{lax}(x) \mathrm{d}x$ isn't elementary and you've only pushed the problem out a little. –  Eric Towers Jun 18 at 6:44
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4 Answers 4

up vote 26 down vote accepted

Elementary functions are finite sums, differences, products, quotients, compositions, and $n$th roots of constants, polynomials, exponentials, logarithms, trig functions, and all of their inverse functions.

The reason they are defined this way is because someone, somewhere thought they were useful. And other people believed him. Why, for example, don't we redefine the integers to include $1/2$? Is this any different than your question about $\mathrm{lax}$ (or rather $\operatorname{erf}(x)$)?

Convention is just that, and nothing more.

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Of course, what's really interesting about Liouville's theorem is the mere idea that it's possible to prove that an antiderivative can't be expressed using a particular family of functions, so the fact that that family is a little arbitrary doesn't detract from the beauty of the result. –  Jack M Jun 16 at 20:55
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I'm not sure that's a fair comparison -- the integers satisfy certain axiomatic properties (e.g., totally-ordered ring whose positive elements are well-ordered) that would no longer be true if you added random rational numbers. I'm not sure there are any similar characteristic properties satisfied by the "elementary" functions. –  user7530 Jun 16 at 20:56
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@JackM: Yes, I think you hit that right on. –  mixedmath Jun 16 at 21:05
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Elementary functions in the sense that Liouville, Chebyshev, Ritt, and others used the term includes any function that is a solution of an algebraic equation with polynomial (or rational function) coefficients. This will include some algebraic functions that are not so-called explicit algebraic functions (i.e. those expressible in finite form as you described). –  Dave L. Renfro Jun 16 at 21:55
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@user7530: If you add $\frac12$ to the integers then all integer multiples of $\frac12$ must also be added. The result is the same as we have done nothing. –  timur Jun 22 at 16:09
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I would approach the question this way. We can think of our "library" of functions being built up recursively: Start with a few basic functions (polynomials, exponentials, logarithms, trig functions, etc.), and start composing, concatenating, integrating, etc. At each stage you have a collection of functions that have been defined "so far". What Liouville's theorem is stating is that:

At any stage, there will be functions whose integrals consist of functions that have not yet been defined.

So for example, you can (if you want) add the function lax $(x)$ to your library and consider it "elementary"... But then as soon as you start to think about the integral of lax $(x)$ you discover that it can't be expressed using elementary functions. So, sure, go ahead and give that function a name and call it elementary; as soon as you do, you realize that its integral can't be expressed using elementary functions. And so on.

Edited to add: A couple of commenters have asked for a reference to the bold-faced paraphrase of Liouville's Theorem above. I should clarify that I don't have a reference for it, and I'm not even sure such a theorem exists (or has been proven). The bold-faced paraphrase is my informal, intuitive interpretation of the meaning of Liouville's Theorem; it seemed to me that was what the OP was looking for. Note that my statement was deliberately vague (it includes "etc." in two different places). I suspect there is probably some more precise refinement of this paraphrase that is true, but I don't know what that refinement would be.

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Do you have a reference for this more general result? I've always suspected something like this might be true, but I've never seen it explicitly stated as a theorem anywhere. –  Jack M Jun 16 at 22:24
    
After looking in a few places, I find no one calling the result in bold "Liouville's theorem". It always refers to elementary functions and fields. Regardless, I am very much interested for a reference to the result in bold. –  RghtHndSd Jun 16 at 23:32
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Not only do I not have a reference for it, I'm not even sure it's true (or been proven). I should clarify that this is how I interpret / think about Liouville's theorem, but it might actually be overstating the truth. (On the other hand I think I've stated it in sufficiently vague form that whatever the truth is, I could probably justify my answer above as being a paraphrase of that truth.) –  mweiss Jun 16 at 23:33
    
This very much intrigues me, so I've made it into a question. –  RghtHndSd Jun 16 at 23:43
    
In fact, after defining $\text{lax}$ as elementary function, you can integrate it in "elementary" functions — in terms of itself. Then the same goes about integral of this integral. –  Ruslan Jun 17 at 12:51
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The motivation here is similar to that in elementary Galois theory where you might study whether or not you can write the roots of a polynomial using the standard arithmetic functions along with the $n$th root function. You might wonder why anyone cares about solving polynomial equations using these restricted functions when you could just use any of a number of numerical methods, or define new functions.

But if you do investigate the restricted problem a large amount of interesting mathematics arises. Solutions to polynomial equations using radicals correspond to solvable groups which are of interest in their own right. This gives plenty of insight into solving more general problems (not to mention that historically this problem helped kickstart the whole field of abstract algebra). The fact that this connection exists now justifies the study of the original restricted problem. And it gives insight into solving wider classes of problems.

If you're writing software to symbolically integrate functions, say, then it'd be risky to ignore Liouville's theorem and Risch's algorithm. They'll give you insight even if you're solving a more general problem.

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I should add that proving an indefinite integral is not one of a given "library" of elementary functions is very, very difficult. Computer algebra systems use the Risch algorithm, hope i have the spelling correct. And you can define any new function you like, but only the ones found widely useful keep such names.

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