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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 $x \in \mathbb{R}$.

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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 $\cos{x} + \{x\}$.

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  • $\begingroup$ 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. $\endgroup$
    – jsmith
    Aug 7 at 22:04
  • $\begingroup$ 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. $\endgroup$
    – jsmith
    Aug 7 at 22:14
  • $\begingroup$ I think it looks good now. $\endgroup$
    – lulu
    Aug 7 at 22:16
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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.

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