Are there "simple" functions whose asymptotic behaviour is impossible to know?

Question

My question is

Does exist a simple function $f \colon A \subseteq \mathbb R \to \mathbb R$ such that it is impossible to know $\lim_{x \to \infty} f(x)$ ?

Of course there are a lot of function whose behaviour is not known; but what happens if we use only elementary functions (the rigourous definition of elementary function involves differential algebra, so let's just imagine a elementary function as a composition of exponentials, logarithms, rational functions and trigonometric functions)?. More clearly the problem is:

Is it possible to create some function (or sequence) "easy to define" for which is not possible to evaluate his asymptotic behaviour?

Edit: with "not possible to evaluate" I mean that the problem to evaluate $$\lim_{x \to +\infty} f(x)$$ is not decidible.

I mean that the limits is impossible to know. We know that $\lim\limits_{n \to \infty} \sin n$ does not exist. — Sewer Keeper, Apr 23 '20 at 16:27
Well, you should say what you mean by "it is impossible to know its asymptotic behaviour". Every function has some asymptotic behaviour, which is described by itself. Are you talking about computability? — Crostul, Apr 23 '20 at 16:32
Fun question. I think the question refers to undecidibility. I would interpret it as: Is there an "easy to define" (i.e., defined in terms of finite composition of elementary function) sequence for which the asymptotic behaviour is undecidable? — fulges, Apr 23 '20 at 16:43
Note that if it is impossible to decide whether there exists $x \in \mathbb R$ for which $f(x) < 0$, then it is impossible to decide whether $x f(x \sin(x))$ is bounded below as $x \to +\infty$. — Robert Israel, Apr 23 '20 at 17:22
@RobertIsrael Does there exist such a function which is moreover elementary? — Sewer Keeper, Apr 23 '20 at 17:52
@SewerKeeper Did you read the statement of the theorem? It explicitly involves only elementary functions. — Kevin Carlson, Apr 23 '20 at 18:39
According to Richardson, yes. Laczkovich showed there exists such a function in the ring generated by the integers, $x$, $\sin(x^n)$ and $\sin(x \sin (x^n))$ for positive integers $n$. — Robert Israel, Apr 23 '20 at 18:43

Oscar Lanzi · Answer 1 · 2020-04-23T17:55:04.710

Consider a sequence defined as follows (also see here):

$a_1=1$

$a_{n+1}\in\{a_n,a_n+1\}$

$a_n=$ number if occurences of $n$ in the sequence.

The recurrence defined above implies

$a_n+1=a_{n+a_{a_n}}$ Eq. 1

Suppose that we assume a leading behavior with the form

$a_n\approx \alpha n^\beta$ Eq. 2

Plugging Eq. 2 into Eq. 1 and applying the binomial power expansion to the right side gives what turns out to be the correct leading behavior:

$\alpha=\phi^{1/\phi^2}$

$\beta=1/\phi$

$\phi=(1+\sqrt5)/2$

This, put into Eq. 2, is indeed the correct leading behavior. But ... the higher order correction terms coming from the binomial expansion are $O(1)$, hence cannot rise above the "noise" generated by the unit jumps in this sequence of whole numbers. Therefore, the leading behavior cannot be expanded into a power series or refined in any other way using analytic functions.

Ah, the glory of detecting a downvote and then trying to guess why. Do tell! — Oscar Lanzi, Apr 23 '20 at 20:43

score 1 · Answer 2 · answered Apr 23 '20 at 17:06

This is an wild attempt.

More than as an answer, it might be thought as a way to get closer to a constructive answer.

Let $P(n)$ be a statement which depends on $n$ with two properties:

for every $n$, it is possible to determine whether $P(n)$ is true or false
the statement "there exists $n_0$ for which $P(n)$ is true for every $n > n_0$" is undecidable

Then define $f : \mathbb{N} \to \mathbb{R}$ by $$ f(n) =\sum_{\substack{m < n , \\ P(m) \text{ is true}}} \frac{1}{m} $$

Then (depending on $P$) I would expect that convergence or divergence of $f(n)$ is undecidable (and tweaking with some signs one can make it into a problem about existence of the limit).

However, this $f$ does not really give an answer to the question without an explicit instance of $P$, because there is an obvious problem if $P(n)$ is true for enough $n$ to have the series diverge, but it is not necessarily true for all $n$.

Hope this helps a bit.

score -1 · Answer 3 · answered Apr 23 '20 at 16:38

-1

Whatever you mean, the following function satisfies your request: $$f(x)= \begin{cases} x & \mbox{ if the Riemann Hypothesis is true}\\ -x & \mbox{ otherwise}\end{cases}$$

Knowing the limit of this function is not impossible, but it's still hard, since it's equivalent on proving the Riemann Hypothesis.

If you want something actually impossible, substitute the Riemann Hypothesis with something impossible to prove, for example the Continuum Hypothesis.

answered Apr 23 '20 at 16:38

Crostul

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But $f(x)$ is not an elementary function, it's not composed of sines, cosines, ... – Sewer Keeper Apr 23 '20 at 16:42
Not even a single value of this function is "known" (in the RH example) or "possible to known" (in the CH) example. A bit unsatisfactory, yet maybe legit – fulges Apr 23 '20 at 16:44
1

This is not even a function, let alone elementary – Raffaele Apr 23 '20 at 16:48
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@fulges Well, $f(0)$ is certainly known – Reveillark Apr 23 '20 at 17:12
Yep..Fair enough :) – fulges Apr 23 '20 at 17:19
@Raffaele in what sense is this not a function? It seems pretty tough for it to take on multiple values! – Kevin Carlson Apr 23 '20 at 18:37

Are there "simple" functions whose asymptotic behaviour is impossible to know?

3 Answers3