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.

Let $f$ be a real non polynomial analytic function. Suppose that the function $f$ assumes arbitrarily large and arbitrarily small values, i.e., for all $K>0$, there are $a,b$ with $f(a)<−K$ and $f(b)>K$.

My question is:

Can we find sufficient and necessary conditions to guarantees this property persists for the derivatives $f^{(k)}, k=1,2,..$.

share|improve this question
1  
Interesting question, +1. By "non polynomial", do you mean a function which needs to be expressed by an infinite amount of terms in its Taylor series? Is it analytic on all of $\Bbb{R}$? In that case, can it be said that the arbitrary large/small values must occur at $\pm \infty$, since otherwise it wouldn't be analytic at the point for which it blows up? Sorry if these questions are basic, I'm not even close to an expert on this. –  Daniel R Sep 14 '13 at 8:16
    
@DanielR: Positive Yes for all your questions. –  DER Sep 14 '13 at 8:23
1  
@EwanDelanoy But can you be sure that none of the derivatives of $f$ only tends to either $+\infty$ or $-\infty$ as $x \to \pm \infty$? (The same thing that happens when deriving $f(x) = x^3$ in the polynomial case.) –  Daniel R Sep 17 '13 at 8:34
1  
@EwanDelanoy How about $f(x) = x^3 + \sin x$ as a counter-example, does it work? –  Daniel R Sep 17 '13 at 8:41
1  
@DanielR Indeed. I understand now. –  Ewan Delanoy Sep 17 '13 at 8:52

1 Answer 1

up vote 1 down vote accepted

This is not a complete answer, but I'll go ahead and post it anyway. Hopefully, it will draw enough attention to correct the probable errors in my reasoning and provide fullblown proofs of the "beliefs" below.

For clarity, I will repeat what was being answered in the comments.

  • The function being "non polynomial" means that it needs to be expressed by an infinite amount of terms in its Taylor series.
  • The fact that it is analytic on all of $\Bbb{R}$ implies that the arbitrarily large/small values must occur at $\pm \infty$. Otherwise it wouldn't be analytic at the point for which it blows up.

Let's start with functions that have well-defined limits (to $\pm \infty$) when $x \to \mp \infty$. Then one and exactly one of the following statements must be true:

$$\lim_{x \to -\infty} f(x) = -\infty\; \text{ and } \lim_{x \to +\infty} f(x) = +\infty \tag{1}$$ $$\lim_{x \to -\infty} f(x) = +\infty\; \text{ and } \lim_{x \to +\infty} f(x) = -\infty \tag{2}$$

In case $f(x)$ is monotone increasing (decreasing), the derivative is positive (negative) for all of $x \in \Bbb{R}$, and therefore the criteria stated in the question are not met.

Belief 1: $f(x)$ is not monotone increasing or decreasing.

My hunch is that any of criteria $(1)$ or $(2)$ is sufficient to make $f(x)$ violate the conditions in the question.

Belief 2: $f(x)$ does not have well-defined limits when $x \to \pm \infty$.

This implies the next belief:

Belief 3: A necessary condition is that $f(x)$ has an oscillation, the magnitude of which grows to infinity as $x \to \pm \infty$.

I'm not sure if this condition is also sufficient (or even necessary for that matter), but I can't come up with any functions that would violate that.


As an example of a real-analytic function that I think have the properties in the question, take $$f(x) = (e^x - e^{-x})\sin kx$$ whose derivative is $$f'(x) = e^{-x} \left[\left(e^{2 x}+1\right) \sin kx+k \left(e^{2 x}-1\right) \cos kx\right]$$

I belive it's real-analytic, since it's built from additions and multiplication of real-analytic functions.

A plot of $f(x)$:

enter image description here

A plot of $f'(x)$:

enter image description here

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.