# Can ANY curve be described by an equation?

1. Jan 24, 2010

### LucasGB

The title says it all: can I pick a piece of paper, draw a completely random curve, and then describe it by an equation? Rephrasing, Can ANY curve be described by an equation?

2. Jan 24, 2010

### Dragonfall

You'll have to define "curve".

3. Jan 24, 2010

### LucasGB

A continuous set of points plotted on a 2D Cartesian coordinate system. Just a 2D graph.

4. Jan 24, 2010

### JSuarez

No, what you have to define is what you mean by equation. In any case, the answer is no: the number of continuous curves is non-denumerable, while the number of (reasonable) ways to describe them is enumerable.

5. Jan 24, 2010

### LucasGB

I see, that's quite interesting! Can you link me to a proof of that statement?

6. Jan 24, 2010

### Phrak

If memory serves me, there is theorem used in artificial neuro networks that states that f(xi) = Sumj gj(xi) may represent an arbitrary function (with some sort of restrictions, perhaps) where j=0..infinity.

In ANN's, the gj may be step functions.

Last edited: Jan 24, 2010
7. Jan 24, 2010

### LCKurtz

You probably already know that the real numbers have a larger cardinality than the rationals by the Cantor argument. The cardinality of the continuous functions on R is at least as large as that of the reals because the map

$\alpha \rightarrow f_\alpha$ where $f_\alpha(x) = (x-\alpha)^2$ is an injection of the reals into the continuous functions on R.

8. Jan 25, 2010

### JSuarez

A link, that I know of, no. Set Theory proves that the cardinality of the set of continuous real function is the same as $$\mathbb R$$ and I know a few books that have the proof. You may check the statement, but not the proof, here:

http://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Sets_with_cardinality_c"

And check the references in that page (Jech's book is very good).

But, anyway, given the above, the argument goes like this: when you describe a function by an equation, whatever the type, you are describing it by a finite string of symbols, over a finite alphabet and the set of all these strings is, at most, enumerable.

Last edited by a moderator: Apr 24, 2017
9. Jan 25, 2010

### CRGreathouse

It's clear that there are uncountably many continuous curves, since there are uncountably many continuous curves of the form f(x) = k for k in R. It's clear that there are countably many formulas over a given alphabet, since they can be listed (order by length and then lexicographically).

10. Jan 25, 2010

### hamster143

The loophole in that argument is that, if we allow formulas with real valued coefficients, their number becomes uncountable (the downside being that our formulas, generally speaking, can't be written down precisely using a finite number of letters, even if their execution involves a finite number of operations). For example, the set of all quadratic polynomials over R is uncountable and has cardinality c, the same as the set of all continuous functions. Therefore, it's possible to construct an isomorphism between the two.

The question we have to ask is whether every continuous function from R to R can be described by a formula that involves a finite number of operations and a finite number of free real coefficients.

11. Jan 25, 2010

### JSuarez

It's not really a loophole, it's a logical question: when you write something like $$a_{2}x^2 +a_{1}x + a_0$$, with $$a_2,a_1,a_0 \in \mathbb R$$, you are not really describing a polynomial (from your post, I see you are aware of that, but you have not drawn the full implication), but a set of polynomials. An expression like the above is called in logic a schema, and denotes a set, not a member of it; the problem is that the parameters are behaving as free variables, not as constants and to describe a particular curve, all constant symbols must refer unambiguously a domain element. In $$\mathbb R$$, you can only refer precisely an enumerable subset.

12. Jan 25, 2010

### Phrak

Why is it being presumed here that functions must be limited to finite number of terms? Am I mistaken?

13. Jan 25, 2010

### DaveC426913

Is there room in here for any curve being approximated to an arbitrary accuracy by an equation?

14. Jan 26, 2010

### mathman

Yes:
I'll assume the domain is [0,1] and the curve is a single valued function. For the nth approximation, evaluate the function at x=k/n for k between 0 and n. A polynomial can be fitted through these points. The approximation gets better as n increases.

15. Jan 26, 2010

### LCKurtz

A nice example is the nth Bernstein polynomial approximation to f on [0,1]:

$$B_n(f)(x) = \sum_{k=0}^n f(\frac k n)\binom n k x^k(1-x)^{n-k}$$

16. Jan 26, 2010

### hamster143

Can we approximate functions in C[0,1] to arbitrary accuracy using formulas with bounded numbers of coefficients?

17. Jan 27, 2010

### LucasGB

Wow, guys, this has got way too advanced for my understanding. What should I take from all of this? That if I establish the condition that the equation must be finite, then there are more curves than equations, and not all curves can be described?

Last edited: Jan 27, 2010
18. Jan 27, 2010

### CRGreathouse

There are more curves than (finite) equations, yes. But this might not be quite what you mean, since there are more real numbers than (finite) equations. In fact, this problem highlights the three infinite cardinalities that come up most frequently:

* There are beth_0 (finite) equations
* There are beth_1 continuous curves
* There are beth_2 curves

19. Jan 28, 2010

### some_dude

I say no. For a random "curve" drawn by hand there exists absolutely no equation describing it (but arbitrarily close approximations). That would be akin to describing the relative position of every single atom you placed on the paper with your writing tool.

20. Jan 29, 2010

### CRGreathouse

Oh no, that's easy. There are only a finite number of atoms on the paper, so that curve can be described by a (parametric) polynomial. :P