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I've been given this exercise (among others) and am trying to verify if it is periodic. For the other problems, I did some graph sketching and solved it but this one is a bit more complicated. At the moment, I just thought about using:

$$\sin(x)=\cos\left(\frac{\pi}{2}-x \right)$$

Yielding:

$$\cos(x)+\sin(\sqrt{3}x)=\cos(x)+\cos \left(\frac{\pi}{2} - \sqrt{3}x \right)$$

And I also tried to use identities such as the ones for $\sin(x) + \cos (y)$ and $\cos(x)+\cos(y)$ without success.

I've managed to find a strategy that could work: We assume that $f(x)=\cos(x)+\sin(\sqrt{3} x)$, if $f$ is periodic, then there is a unique smallest positive number $t$ such that for all $x$, we have:

$$f(x)=f(x+t)$$

All we need to do now is pick $x,x'$ such that finding solutions for $t$ in

$$f(x)=f(x+t)\\f(x')=f(x'+t)$$

yield different results. One of the "strategic" values is $x=0$, we then have:

$$1=\cos(t)+\sin(\sqrt{3}t)$$

Now if we come up with a certain $x'$ such that

$$\cos(x')+\sin(\sqrt{3}x')=\cos(x'+t)+\sin(\sqrt{3}(x'+t))$$

implies that $t$ needs to be different in both equations, I guess this shows it is non-periodic. The trouble is that finding such an $x'$ doesn't seem easy.

Red Banana
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    If the function were periodic, $\sqrt3$ would be a rational number. – user May 19 '20 at 19:24
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    The period for $\cos x$ is $2\pi$ and the period for $\sin(\sqrt{3}x)$ is $2\pi/\sqrt{3}$. If such function is periodic, then such period, say $T$, will be a common integer multiple of both two numbers. However, this is impossible. –  May 19 '20 at 19:24
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    Instead of looking at them together, think about each function separately. If $f$ and $g$ are each periodic with period $p_1$ and $p_2$, respectively, what must be true about $p_1$ and $p_2$ in order for $f+g$ to be periodic? – Clayton May 19 '20 at 19:25
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    This broader question is interesting to look at : https://math.stackexchange.com/q/661638/399263 – zwim May 19 '20 at 19:27
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    If the function were periodic, an integer multiple of $2 \pi $ would equal an integer multiple of $\sqrt 3$ which woud mean that $\pi^2$ would be rational, which is not true, because $\pi$ is transcendental over the rationals. – P. Lawrence May 19 '20 at 19:38
  • @Clayton you should give some additional limitations of your functions, for example continuity and boundness. – Pavel Kozlov May 19 '20 at 20:06

3 Answers3

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Note $$\cos(x)+\sin(\sqrt{3}x)=2\cos(\frac{\sqrt3+1}2x-\frac\pi4) \cos(\frac{\sqrt3-1}2x-\frac\pi4) $$ For periodicity $\omega $, the following shall hold

$$\frac{ \sqrt3+1}2 w = 2\pi m ,\>\>\>\>\>\frac{ \sqrt3-1}2 w = 2\pi n $$ or,

$$\frac nm = \frac{\sqrt3-1}{\sqrt3+1}$$

which is impossible, hence no periodity.

Quanto
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Derivatives of a periodic function also have the same period. Muliples and linear combinations of functions with the same period also have the same period. In this case, differentiating twice, this means that if $$\cos(x) + \sin\left(\sqrt3 x\right)$$ is periodic with period $T$ then so is $$\cos(x) + 3 \sin\left(\sqrt3 x\right).$$ And then, taking appropriate linear combinations, also $\cos(x)$ and $\sin \left(\sqrt3 x\right)$ both have period $T$. This only leaves $T=0$ (if you would call that a period) as an integral multiple of both $2\pi$ and $2\pi/\sqrt3$.

WimC
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You can choose $k,l\in \mathbb{Z}$ such that $|4k\sqrt{3}-(4l+1)|$ is less than a given $\epsilon>0$, so for $x_0=2\pi k$ the function $f(x_0)$ is more than $2-\epsilon$. If $T$ is a period of $f$, then $f(x_0+nT)=f(x_0)$ for every integer $n$, so $\cos nT = \cos (x_0+nT)$ is more than $1-\epsilon$. It is true only if $T=2\pi l$ for some integer $l$. As it is period of $\cos x$ to, $T$ is also a period of $\sin \sqrt{3}x$. It is wrong because $\sqrt{3}$ is irrational.

Pavel Kozlov
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