2
$\begingroup$

Is there any real-valued function $f(x)$ of a real variable $x$ with $n^{th}$ and $m^{th}$ derivatives never equal for nonequal nonnegative $m$ and $n$ and where the $n^{th}$ derivative of $f$ never has a vertical asymptote?

The idea is that f should be infinitely differentiable like $e^x$ or $sin(x)$ without being periodic by differentiation. In this case I would call $e^x$ 1-periodic by differentiation and $sin(x)$ 4-periodic by differentiation.

The asymptote requirement is because the solution $f(x)=\frac{1}{x}$ is obvious.

Also, is there a name for this class of functions?

$\endgroup$
3
$\begingroup$

$f(x)=\frac{1}{1+x^2}$ should do the trick.

$\endgroup$
  • $\begingroup$ Yes, it does. Now I am wondering if there is a name for these kinds of functions. $\endgroup$ – mikebolt Feb 25 '14 at 3:27
  • 3
    $\begingroup$ Having periodic derivatives is actually the much rarer case; it's just a rarer case that a lot of the standard elementary functions happen to fall into. $\endgroup$ – Micah Feb 25 '14 at 4:03
  • 2
    $\begingroup$ Having derivatives is a rare case that many functions we study fall into. $\endgroup$ – Ross Millikan Feb 25 '14 at 4:22
1
$\begingroup$

I like $xe^x{}{}{}{}{}{}{}{}$.

$\endgroup$
1
$\begingroup$

$e^{2x}$ is not so bad either.

$\endgroup$
  • $\begingroup$ Right, but that's no so interesting. You just end up with a scalar multiple of the function. $\endgroup$ – mikebolt Feb 25 '14 at 3:47
  • $\begingroup$ I guess you like $e^{x^2}$ better. $\endgroup$ – LeoTheKub Feb 25 '14 at 3:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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