Can ANY curve be described by an equation?

[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?

 

[Dragonfall]

You'll have to define "curve".

 

[LucasGB]

You'll have to define "curve".

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

 

[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.

 

[LucasGB]

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.

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

 

[Phrak]

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

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

 

[LCKurtz]

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

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.

 

[JSuarez]

Can you link me to a proof of that statement?

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.

 

[CRGreathouse]

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.

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

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).

 

[hamster143]

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).

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.

 

[JSuarez]

The loophole in that argument is that, if we allow formulas with real valued coefficients, their number becomes uncountable

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.

 

[Phrak]

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

 

[DaveC426913]

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

 

[mathman]

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

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.

 

[LCKurtz]

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.

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

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

 

[hamster143]

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

 

[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?

 

[CRGreathouse]

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

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

 

[some_dude]

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?

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.

 

[CRGreathouse]

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.

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

 

[some_dude]

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

Haha, okay I stand corrected. (If i knew a little more, i'd have some rebuttle referencing the Heisenberg uncertainty principle :P)

 

[CRGreathouse]

Haha, okay I stand corrected. (If i knew a little more, i'd have some rebuttle referencing the Heisenberg uncertainty principle :P)

Hey, if all you need to know is position, the Uncertainty Principle is no problem. :p

 

[some_dude]

Hey, if all you need to know is position, the Uncertainty Principle is no problem. :p

Right. I swear I knew that.

 

[EnumaElish]

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

Yes; there is a theorem that says that the set of polynomials is a dense subset of the set of continuous functions on any closed interval. This is similar to the rationals (countable) being a dense subset of the reals (uncountable).