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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?
Partial answer.
- 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 $h=g^{-1}\circ f$.
- $f$ 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 $\frac{\sin(x)}{e^x}\cdot e^x$.
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.
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.
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.
