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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?

3 Answers 3

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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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    Your example is not differentiable. Nor is the rest of the answer a proof. – 2018-05-13
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    The proof can be arrived using fourier series i guess you research it – 2018-05-13
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I'm interested in this question too. The example given using the nearest integer function will not suffice because the nearest integer function can be expressed using inverse trigs. I too would like a proof.

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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.