Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Do they necessarily have asymptotes? Can they be finite over the first interval ($0$ to $x$), infinite over the second ($x$ to $y$), and return to be finite over a third ($y$ to $z$)? When expressed as an infinite series, do the endpoints of their radii of convergence correspond to asymptotes? Thanks.

share|improve this question
1  
It is unclear which functions you are considering. The study of function spaces of various kinds, and the correct definition of a function space to accommodate mathematical concepts (e.g. fourier transform) is a huge area. Bear in mind that in a technical sense "most" continuous functions are not differentiable, and have properties which are not possessed by familiar elementary functions and which seem weird at first glance ... –  Mark Bennet Mar 26 '12 at 7:17

1 Answer 1

up vote 1 down vote accepted

"Closed form" is an arbitrary term which basically means nothing.

In its most extreme form, it would mean that only polynomials are in "closed form". Common wisdom would probably add trigonometric and exponential functions into the mix, but there is no substantial reasons why these functions would be considered "good" while others are "bad".

To illustrate what I mean, consider these two functions: $$ f(x)=\sum_{k=0}^\infty\frac{x^k}{k!},\ \ \ g(x)=\sum_{k=0}^\infty\frac{x^k}{(5k+1)!}. $$ The first one is the exponential, a very respected function with its own notation, i.e. $f(x)=e^x$. The second function is "exotic" (I guess, I didn't make an effort to think about it, the point is just that it isn't one of the canonical functions) but it is probably almost as "good" as the exponential.

Finally, to address your question directly, given almost any possible property of a function ("good" or "bad" behaviour), there is most likely a "non-closed-form" function with that property.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.