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By "curved", I mean that there is no portion of a graph of a function that is a straight line. (Let us first limit the case into 2-dimensional cases.. and if anyone can explain the cases in more than 2-dimensional scenarios I would appreciate.. but what I really want to know is the 2-dimensional case.)

Let us also say that a function is surjective (domain-range relationship).

So, can every continuous and curved function be described as a formula, such as sines, cosines, combinations of them, or whatever formula?

thanks.

3 Answers 3

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No, no matter what definition of formula you use. In fact, almost all "curved" continuous functions cannot be expressed by a formula. This is because a formula is a finite length sequence of symbols chosen from a finite alphabet, and so there are only countably many formulas. Yet there are uncountably many "curved" functions (for example, $x^2-r$ for any $r\in\mathbb R$), so most of these cannot be expressed by a formula. Note that just writing $x^2-r$ isn't a formula until I've given some formula for the particular value of $r$, which is not possible for most $r\in\mathbb R$.

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No. First, because when you say "whatever formula" you probably include the elementary operations, maybe fractional power, and a few known functions (sine, cosine, exponential, say); there are many other functions, not only continuous but infinitely differentiable that are known to be impossible to express in terms of those elementary functions mentioned above.

But there is also something deeper, which is that when you say "continuous", you are graphically thinking of a differentiable function. Any function that is expressed "by a formula" is almost surely differentiable, and most likely infinitely differentiable. But there are continuous functions which are not even differentiable at a single point.

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The answer depends a little on what you consider a formula. The function $f(x)=-x^2$ for $x<0$, $f(x)=x^3$ for $x\ge 0$ is nicely curvy, and continuous everywhere. The function is also surjective, and injective.