0
$\begingroup$

We know that the volume of a cube can be represented by the function: $V(x)=x^3$, where $x$ is side length. $x^2$ can represent the volume of some material that has a constant side ($1$). The function $(x-1)(x+2)$ can represent an area that depends only on $x$. You get the point.

So my question is:

Can any algebraic or transcendental function (or combination) be the mathematical model that describes some process in the real world?

For example, the function $x^2$ describes the behavior of an area as a function of $x$.

What does a more complex function like this describes: $\displaystyle\frac{x-1}{(x+2)(x^2+4)}$ ?

We often do: real process $\to$ formulate equation that describes it. But can we do it the other way around?

EDIT:

As pointed out in the comments, maybe this question describes better what I mean:

Can all functions be used to create a mathematical model that describes something that we consider 'not mathematical'?

$\endgroup$
4
  • 1
    $\begingroup$ Your question essentially boils down to, "Can all functions be used to create a mathematical model that describes something that we consider 'not mathematical'?" (I considered saying "physical," but there's also "real" things that are arguably not physical, like thoughts and feelings.) I think that raises way too many philosophical questions, like, "What is and is not mathematical?" On top of that, I seriously doubt either a "yes" or "no" answer could be proven correct, since there is a vast, vast amount of unknown that may or may not be described using mathematical constructs. $\endgroup$
    – jpmc26
    Commented Jul 7, 2015 at 22:37
  • $\begingroup$ @jpmc26 That's what I meant. $\endgroup$
    – gab06
    Commented Jul 7, 2015 at 23:02
  • $\begingroup$ I like @RobertIsrael's answer. I would suggest that every real word scenario can be described with words. Even if we allow infinitely long descriptions, this is still a countable number of scenarios. Even if we allow for a countable vocabulary constructed from a finite alphabet, the set of all such descriptions is of cardinality $\aleph_1$. But the set of all functions $\mathbb R\to\mathbb R$ has cardinality $\aleph_2$. $\endgroup$
    – Plutoro
    Commented Jul 7, 2015 at 23:38
  • $\begingroup$ @AlexS No need to assume the generalized continuum hypothesis here. The set of all descriptions has cardinality $\le c = 2^{\aleph_0}$. The set of functions $\mathbb R \to \mathbb R$ has cardinality $2^c$. These are not necessarily $\aleph_1$ and $\aleph_2$. $\endgroup$
    – Robert Israel
    Commented Jul 8, 2015 at 0:17

2 Answers 2

Reset to default
12
$\begingroup$

Functions in the modern sense of the word are much more general than the expressions that you seem to think of as "functions". In particular, most functions (in the sense of cardinality) cannot be specified in a finite number of symbols in a given language, and therefore would not be able to represent anything you could name, whether in the real world or out of it.

$\endgroup$
5
$\begingroup$

It's difficult to say what you mean by the existence of a real world application to each function. Certainly, there could be a very contrived application, but I think you would be hard pressed to find a natural application to some of these functions that mathematicians find useful.

The Dirichlet function: $$f(x)=\cases{0 & \text{ if $x$ is irrational}\\ 1 & \text{ if $x$ is rational.}}$$

Thomae's function: $$f(x)=\cases{0 & \text{ if $x$ is irrational}\\ \frac{1}{q} & \text{ if $x=\frac{p}{q}$ where $p$ and $q$ are relatively prime integers.}}$$

Weierstrass function: a continuous function which cannot be said to have a slope at any point (it is nowhere differentiable).

Peano curve: any parametric function $f:\mathbb R\to\mathbb R^n$ which is surjective (i.e., it covers every point in $\mathbb R^n$).

$\endgroup$
2
  • $\begingroup$ Thomae's function looks intriguing. As far as I can tell it is continuous for all irrational $x$ and discontinuous for all rational $x$. $\endgroup$
    – kasperd
    Commented Jul 8, 2015 at 7:34
  • $\begingroup$ Yes, that's exactly it. $\endgroup$
    – Plutoro
    Commented Jul 8, 2015 at 12:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .