# Proving that $\cos{x} + \{x\}$ is not periodic?

Let's say I have the function $$\cos{x} + \{x\}$$, where $$\{x\} = x - \lfloor x \rfloor$$. How would I prove that the function is not periodic? It seems intuitively true to me, but I can't seem to be able rigorously prove it. I.e., I can't negate the possibility that there is some positive $$p$$ such that $$\cos{x} + \{x\} = \cos{(x + p)} + \{x + p\}$$ for all $$x \in \mathbb{R}$$.

We can pick two or more convenient values of $$x$$ that lead to a contradiction. If we pick $$x = 0$$, then $$\cos(0) + \{0\} = 1 = \cos{p} + \{p\}$$.

If we use $$x = \pi$$, we have $$-1 + \{\pi\} = -\cos{p} + \{\pi + p\}$$. Adding this with the equation we got from the $$x = 0$$ case, we get that $$\{ \pi\} = \{p\} + \{ \pi + p \}$$. In other words, $$\pi - \lfloor \pi \rfloor = p - \lfloor p\rfloor + \pi + p - \lfloor \pi + p \rfloor$$, which simplifies to $$0 = 2p + \lfloor \pi \rfloor - \lfloor p \rfloor - \lfloor \pi + p \rfloor$$. Thus, $$p$$ must equal to some integer divided by two. If that integer is even, then we reach a contradiction, as we will prove that $$p$$ cannot be a positive integer. We will also prove that $$p$$ is not some number with fractional part equal to $$0.5$$.

If $$p$$ is a positive integer, then from $$\cos{x} + \{x\} = \cos{(x + p)} + \{x + p\}$$, we get $$\cos{x} = \cos{(x + p)}$$ This can't be the case, as any value $$p$$ that satisfies this for all $$\mathbb{R}$$ on a continuous function must be an integer multiple of the fundamental period, which is $$2 \pi$$ (an irrational number). (citation, though I can't seem to find the proof for this statement. If anybody can share it either as a comment or as a further answer,that would be great).

If $$\{p\} = 0.5$$, then again from $$x = 0$$, we get that $$\cos{p} = 0.5$$. However, the only values of $$p$$ such that $$\cos$$ is equal to $$1/2$$ are irrational, a contradiction.

Therefore, there is no value of $$p$$ that satisfies a period for $$\cos{x} + \{x\}$$.

• It solves the problem because (i) it shows that every period has to have fractional part of either 0 or 0.5 and (ii) rules out those two possibilities. Aug 7 at 22:04
• Fair enough. To be honest, I wrote the paragraphs in the order in which I was trying out things to figure out the solution which is why it's a little bit out of order haha. Thanks for your suggestion! I'll try to implement your writing advice. Aug 7 at 22:14
• I think it looks good now.
– lulu
Aug 7 at 22:16

Define $$f(x):= \{x\};\ g(x) := \cos(x).$$ Suppose $$f+g$$ has period $$\alpha.$$

Then,

$$f(-\alpha) + g(-\alpha) = f(0) + g(0) = 1 = f(\alpha) + g(\alpha).\quad \text{This gives:}$$

$$\{-\alpha\} + \require{cancel} \cancel{\cos(-\alpha)} = \{\alpha\} + \require{cancel} \cancel{\cos(\alpha)}$$

$$\implies \alpha = 0.5 + k,\ k\in\mathbb{Z}.$$ Therefore,

$$f(0) + g(0) = f(2(0.5+k)) + g(2(0.5+k)),$$

i.e.

$$1 = 0 + g(1+2k),\ k\in\mathbb{Z},$$

which is false.