1
$\begingroup$

Let $C,D > 0$. We call a function $f : \Bbb R \to \Bbb R$ pretty if $f$ is a $\Bbb C^2$-class, $|x^3 f(x)| \leq C$ and $|xf''(x)| \leq D$.

(i). Show that if $f$ is pretty, then, given $\epsilon > 0$, there is a $x_o \geq 0$ such that for every $x$ with $|x| \geq x_o$, we have $|x^2 f'(x)| < \sqrt{2CD} + \epsilon.$

(ii). Show that if $0 < E < \sqrt{2CD}$ then there is a pretty function $f$ such that for every $x_o \geq 0$ there is $a x > x_o$ such that $|x^2 f'(x)| > E$.

This is a problem from Brazilian undergraduate math olympiad. I am not interested in solutions. What I'd like to know is where this definition came from. Does somebody know if it is related to something more advanced, or was it created just for the problem? If it is the latter, I'd like to know how this problem was created. What was the thought process? What was the inspiration? Is it possible to know?

$\endgroup$
  • 1
    $\begingroup$ Hello! It would be helpful to those answering your question to format your question using MathJaX. $\endgroup$ – rb612 May 20 at 0:10
  • 1
    $\begingroup$ Definitions like this are usually not mathematically useful, and are rather typically created for such exams. $\endgroup$ – The Count May 20 at 0:28
1
$\begingroup$

It's just a prefabricated definition, it was written that way so that (i) and (ii) hold, it has no other use than to evaluate the skills of the participants.

$\endgroup$
  • $\begingroup$ Do you have any idea about how the author of the problem created this "definition" ? I have never seen anything similar and can't imagine how someone could think of it. $\endgroup$ – danilocn94 May 20 at 1:36
  • 1
    $\begingroup$ Are you familiar with sequences epsilon limit definition? If you are, you might know that when calulating the M for which if n>M |xn-L|<epsilon, you start by developing your thesis and then you get to a condition that if satisfied, makes your thesis work. In other words, you start by what you want to prove, you develop it until you get basic conditions, for example prove that x>2 if -x-6<-8, assume the thesis x>2, so x+6>8 and -x-6<-8, there's your condition, of course a problem such as your would require an extensive proof but that's the spirit of how you get those crazy conditions. $\endgroup$ – Mario Aldean May 20 at 2:35

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.