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Let $f(x)=x-[x]$ and $g(x)=\tan x$.

How could we see that $f(x)-g(x)$ is not a periodic function?

This will show that the sum of two periodic functions need not be a periodic function.

I hope the answer has enough details so that I could catch you.

amWhy
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Paul
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  • Let $T_1$ be the period of $f$ and $T_2$ be that of $g$ then $f+g$ is periodic iff $\frac{T_1}{T_2}\in\mathbb{Q}$ – user5402 Jul 27 '13 at 17:09
  • @metacompactness Note that it is possible for a (highly discontinuous) function to be periodic with no well-defined minimal period. For instance $f$ could be periodic with period $1$ but also have period $\pi$. In this case $f+f$ is periodic even though $1/\pi \notin \mathbb Q$. I admit this is a pathological example and that there is no canonical period $T$ for $f$, but I think some care needs to be taken not to make unwarranted assumptions. – Erick Wong Jul 27 '13 at 17:22
  • "periodic with no well-defined minimal period", what does that mean? can you give an example? Of course we're studying functions $f:~I\subset\mathbb{R}\longrightarrow J\subset\mathbb{R}$ – user5402 Jul 27 '13 at 18:07
  • @metacompactness Yes, here is an example: use the axiom of choice to choose a unique representative from every coset of the additive subgroup $\langle 1,\pi \rangle$ in $\mathbb R$. Then define $f(x)$ to be the representative of the coset containing $x$. Now $f(x+1) = f(x)$ and $f(x+\pi) = f(x)$. In fact, the periods of $f$ are precisely the numbers of the form $m\pi + n$ with $m,n\in\mathbb Z$, and there is no minimal positive element of this form. – Erick Wong Jul 27 '13 at 18:12
  • @ErickWong Is $f$ an application? for example what is the value of $f(\pi)$? – user5402 Jul 27 '13 at 20:07
  • @metacompactness Since it is an AC-dependent construction, it's hard to give an explicit basis. However, one could easily take $f(0)=f(\pi)=f(1)=0$. The idea is that there exists a set $S\subseteq \mathbb R$ such that every $x\in \mathbb R$ can be written uniquely as $x = s + m\pi + n$ for some $s \in S$, $m,n \in \mathbb Z$. Then set $f(x) = s$. – Erick Wong Jul 27 '13 at 22:56
  • I just noticed Qiaochu Yuan's pretty construction here: http://math.stackexchange.com/questions/1079/sum-of-two-periodic-functions. This example compellingly shows why the proofs relying only on $1,\pi$ not having a common multiple are fundamentally incomplete. – Erick Wong Dec 24 '13 at 00:23

3 Answers3

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First, notice the range of g is not $\mathbb{R}$ but $\mathbb{R} \cup \{ \infty \}$. Second, $g$ and hence $f - g$ take the value $\infty$ $\color{red}{\text{at and only at}}$ $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm\frac{5\pi}{2}, \ldots$. This means if $f - g$ is a periodic function, then its period must have the form $n\pi$ where $n \in \mathbb{Z}_{+}$. If $n\pi$ is a period, we will have:

$$f(n\pi) - g(n\pi) = f(0) - g(0)\quad\implies\quad n\pi - \lfloor n\pi\rfloor = 0 \quad\implies\quad \pi = \frac{\lfloor n\pi\rfloor}{n} \in \mathbb{Q}$$

This contradicts with the known fact that $\pi$ is an irrational number.

achille hui
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4

For example, $sin(x) + sin(\sqrt{2} x)$ is not periodic

Maxim Umansky
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    It doesn't look like it is periodic, but what is the proof that it isn't? – Pete L. Clark Feb 19 '14 at 05:51
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    Let $f(x) = \sin(x) + \sin(\sqrt{2} x)$. Then $2f(x) + f''(x) = \sin(x)$ and $f(x) + f''(x) = -\sin(\sqrt{2}x)$. If $f$ has period $p > 0$, then $2f + f''$ and $f + f''$ also have period $p$. The first statement implies that $p$ is an integer multiple of $2\pi$, and the second implies that $p$ is an integer multiple of $2\pi\sqrt{2}$. But this is impossible because $\sqrt2$ is irrational. – Dave Radcliffe Jul 30 '18 at 00:34
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As user84559 said above, the period T of the difference must be an integer multiple of $f(x)$ and $g(x)$. Now the period of $f(x)$ is $1$, and the period of $f(x)$ is $pi$. So the period could be integer multiples of each period, such as $1, 2, 3, ...$ or $pi, 2pi, 3pi, ... $However, there are no common multiples of $1$ and pi. In other words any of $1, 2, 3,...$ will never equal $pi, 2pi, 3pi...$

Ovi
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  • Is it (rigorously) obvious that this is a necessary condition to be a period of $f+g$? One can of course contrive periodic functions $f$, $g$ so that the period of $f+g$ is not a multiple of the respective periods. I find myself exploiting salient features of this specific pair of functions rather than applying a general principle. – Erick Wong Jun 03 '13 at 01:38
  • Yeah your approach would be better and there would be more to learn from it but this is the first thing that came to mind and I didn't have a lot of time when I wrote this answer. – Ovi Jun 03 '13 at 03:44
  • @ErickWong: Actually, what would be 2 functions f and g such that f+g has a period which is not a multiple of the respective periods? – Ovi Jun 03 '13 at 05:17
  • $f(x) = \sin x$ and $g(x) = -\sin x$. Here $f+g$ has many periods which are not multiples of $2\pi$. Similarly one could choose $f+g$ to have period $\frac{\pi}{23}$ while $f$ and $g$ have period $2\pi$ (choose $f+g$ first, then $f$). – Erick Wong Jun 03 '13 at 06:10
  • But you have to prove that there exist such f and g that they have a period of 2pi and that f+g had period pi/23. Otherwise, you would need a specific example. – Ovi Jun 03 '13 at 06:21
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    Huh? I just told you how to find an example: choose $f+g$ first, then $f$. For instance take $f(x)+g(x) = \sin(46x)$ and $f(x) = \sin(x)$. – Erick Wong Jun 03 '13 at 06:23
  • @ErickWong $f(x)=\sin x$ and $g(x)=-\sin x$ have a period $2\pi$ and their sum $0$ has a period $0$ which is a multiple f $2\pi$. The same applies to your 2nd example. – user5402 Jul 27 '13 at 17:05
  • @metacompactness $0$ is a trivial period of any function, so I'm not sure what your point is. Since we are talking about periods in the general sense (rather than minimal periods), the statements "$f$ has a period which is a multiple of $x$" and "$f$ has a period which isn't a multiple of $x$" are entirely consistent. – Erick Wong Jul 27 '13 at 17:18
  • @ErickWong by period I mean the prime period and remember we're studying functions $f:~I\subset\mathbb{R}\longrightarrow J\subset\mathbb{R}$ whether continuous or discontinuous. – user5402 Jul 27 '13 at 18:08
  • @metacompactness From the discussion in the main question, it is not always possible to define the period of a periodic function $f$ in the general setting of (partial) functions on $\mathbb R$. My original point was that even when the period of $f+g$ is uniquely defined, it's completely false that this unique period must be a common multiple of $f$'s and $g$'s periods. – Erick Wong Jul 27 '13 at 23:10