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Hypothesis:

Given a non constant differentiable periodic function $f(x),\ \mathbb{R}\to\mathbb{R}$, which has an explicit finite form involving only elementary functions. The expression for $f(x)$ must include trigonometric functions, or equivalently, complex exponents.

In other words, we can't construct a periodic function from only roots, real exponents, logarithms and polynomials.

Is there a simple proof or counter example for the above hypothesis?

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  • $\begingroup$ It seems like it must be true but I'd imagine it would be quite tedious to prove. You'd want to show that all compositions of $x^n,e^x,\ln(x),\ldots$ are non-periodic but I can't imagine many simple ways of dealing with it other than going through each case. Maybe we could potentially try a set of inductive arguments (for each type of elementary function) that it either explodes or tends to a constant. At the very least, I see this needing about $7$ lemmas for each of $x^n, e^x,\ln(x),\ldots$. $\endgroup$
    – Jam
    Jun 5, 2019 at 21:35
  • $\begingroup$ Also, out of interest, why did you include the stipulation of differentiability? The elementary functions are closed under differentiation, so surely that's redundant? $\endgroup$
    – Jam
    Jun 5, 2019 at 21:49
  • $\begingroup$ @Jam - I'll agree that the elementary functions plus the finite requirement are enough. $\endgroup$
    – nbubis
    Jun 6, 2019 at 9:36
  • $\begingroup$ Let $f_0(x)=x$ and for each $n$, $f_n(x) = H_n(f_0(x),\ldots,f_{n-1}(x))$ where $H_n$ is one of the operations allowed in the elementary functions (constant, addition, multiplication, exponentiation, $\log$, differentiation). Your elementary function is $f_N : \Bbb{R} \to \Bbb{R}$. If no $e^z$ appears in the $H_m, m \le n$ or if the imaginary part of the argument of $e^z$ is constant then $f_n$ is constant or $f_n^{-1}(a)$ is finite for every $a\in \Bbb{R}$ so $f_n$ can't be periodic. $\endgroup$
    – reuns
    Jun 6, 2019 at 17:02
  • $\begingroup$ @reuns This seems like a good answer but I'm a bit confused by it. What if we took $f_n=\frac{1}{x^2+1}$? It's neither constant nor has a finite inverse (since $f_n^{-1}(0)=\pm\infty$) but isn't periodic. $\endgroup$
    – Jam
    Jun 6, 2019 at 17:45

4 Answers 4

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Partial answer.

  1. If $f$ is a composition, then the last composed functions can be ignored if they are injective. Thus, for anything of the form $e^{h},h^{2n+1},\ln h,\ldots$, we need just consider $h(x)$.

Proof: Suppose we were to construct a periodic, composite function $f(x)=g\circ h(x)$, where $g$ is injective and thus aperiodic. Then $h$ must have already been periodic since it is trivially $h=g^{-1}\circ f$.

  1. $f$ cannot be a rational function with an injective function as its variable.

Proof: If $f(x)$ were a periodic polynomial then, for some $x_0$, $f(x)-f(x_0)$ would be a polynomial with infinite roots, $x_0+nP$, where $P$ is a period and $n\in\mathbb{Z}$. This would violate the fundamental theorem of algebra. This can be generalised to any rational function $r(x)=\frac{a(x)}{b(x)}$ since the polynomial $a(x)-r(x_0)b(x)$ would have infinite roots, $x_0+nP$. We can then generalise it further to $r\circ h(x)$ for any rational function, $r$, with an injective function, $h$, as its variable since it would have at most as many roots as $r(x)$. So all functions of the form $\frac{e^{2x}-3e^x}{e^{2x}+1},\frac{\ln(x)^2+1}{\ln(x)},\ldots$ cannot be periodic.

The question remains for combinations of different elementary functions such as $\frac{e^x+\ln(x)+\ldots}{x+\ldots}$. A difficulty in the problem is that the composition of two aperiodic, elementary functions can form a periodic function. For example, take $g(x)=\cos\sqrt{x}$ and $h(x)=x^2$. So too can the sum or product of two aperiodic, elementary functions. For instance, take $\left(\sin x+e^x\right)+\left(-e^x\right)$ or $\frac{\sin(x)}{e^x}\cdot e^x$.

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    $\begingroup$ An excellent start. $\endgroup$
    – nbubis
    Jun 7, 2019 at 6:26
  • $\begingroup$ Notes: We might expect $f$ to be $\mathcal{O}(1)$, like $\sin x$ and $\cos x$ but alas, this wouldn't be the case for discontinuous periodic functions like $\tan x$. At the very least, it's apparent if $f$ tends to $+\infty$, it must do so on a bounded interval, otherwise it would be identically $+\infty$ everywhere. So $f$ also can't be a function explodes to $+\infty$ without discontinuities, e.g. $\frac{e^x}{x^2+1}$. $\endgroup$
    – Jam
    Jun 7, 2019 at 10:57
  • $\begingroup$ The tricky cases are ones like $f=\frac{a(x^2)}{b(x^2)}$ for $a(x)=e^x-3x^2+\frac{1}{x^2+1},\,b(x)=e^x$ since they are $\mathcal{O}(1)$ but it's not immediately obvious how many roots the numerator has - particularly if it's a heavily nested function. Also, we know that $f$ must not have a definable inverse at $\pm\infty$. Lastly, I think the best tactic would be to show that any non-constant function formed by successive addition, multiplication, exp and logs has only finitely many real roots (up to a constant). Despite seeming obvious, I can't think how to prove it. $\endgroup$
    – Jam
    Jun 7, 2019 at 11:30
  • $\begingroup$ On second thoughts, that tactic doesn't work. Take $f=\sqrt{x^2}+\sqrt{\left(x-1\right)^2}-1$, which has infinitely many roots: $x\in[0,1]$. $\endgroup$
    – Jam
    Jun 7, 2019 at 11:47
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Yes. You are right. But I would like to propose a nearly-counterexample (which of course does not count, but still).

If you count the Bernoulli polynomials of negative order, then:

$$B_{-1}(x)+B_{-1}(1-x)=\psi^{(1)}(x)+\psi^{(1)}(1-x)=\pi^2 \csc^2 \pi x$$

This is a periodic function. And elementary. The only problem is, the Bernoulli polynomials of negative order are not polynomials or even elementary functions. But like Bernoulli polynomials of positive order they are cases of Hurwitz Zeta function.

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No. For if you define $f:\mathrm R\to \mathrm R$ by the finite expression $$2x-x^2$$ for $$x\in [0,2]$$ and $$x^2-6x+8$$ for $$x\in [2,4],$$ and elsewhere by $$f(x)=f(x\pm 4),$$ then we see that this function is non-constant, differentiable, and periodic, yet contains no circular functions.

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    $\begingroup$ Nice try, but I don't count an 'if' statement as an elementary function. $\endgroup$
    – nbubis
    Jun 1, 2019 at 17:31
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If we consider $[x]$ to be elementary then $x-[x]$ is also periodic with period $=1$
Other than that, Only trigonometric functions are periodic as they are circular functions which repeat after an interval and all periodic functions are composed of trigonometric functions in some form or other.

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    $\begingroup$ Your example is not differentiable. Nor is the rest of the answer a proof. $\endgroup$
    – nbubis
    May 13, 2018 at 8:42
  • $\begingroup$ The proof can be arrived using fourier series i guess you research it $\endgroup$ May 13, 2018 at 8:43

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