# $\cos(x)+\cos(x\sqrt{2})$ is not periodic

Show that the function $$f(x)=\cos(x)+\cos(x\sqrt{2})$$ is not periodic. I tried $x = a$ and $a\sqrt{2}$. I am guessing that the method of contradiction would be of some help over here. What else should I try?

• To prove it isn't periodic you have to prove there is no $\alpha$ such that for all $x$ you have $f(x) = f(x + \alpha)$. If you prove this can't work for one particular $x$, you are set. – vonbrand May 13 '13 at 11:00
• @vonbrand:I know that.Anyways thanks! – Shaswata May 13 '13 at 11:04
• Please explain what your thoughts are, otherwise it is hard to give useful suggestions. – vonbrand May 13 '13 at 11:05
• What about try $x = 0$? – mez May 13 '13 at 11:08

Assume $\cos(x) + \cos(\sqrt{2} x) = \cos(x+T) + \cos(\sqrt{2} x+ \sqrt{2} T)$.

Let $x=0$. Then

$$2=\cos T + \cos \sqrt{2}T$$

Since the cosine is at most one, this means that simultaneously $\cos T = 1$ and $\cos \sqrt{2} T = 1$. This is equivalent to:

$$T = 2\pi n, \ \mbox{for some } n \in \mathbb{Z}$$ $$\sqrt{2}T = 2\pi m, \ \mbox{for some } m \in \mathbb{Z}$$

Substitute $T$ from the first into the second:

$$\sqrt{2} (2\pi n)= 2\pi m$$ $$\sqrt{2}=\frac{m}{n}$$

So $\sqrt{2} \in \mathbb{Q}$ which is a contradiction, or $T=0$. Either way, you're done.

There's only one value of $x$ for which $f(x)=2$. Proof: Since $\cos\phi\le1$ this is only possible, if $\cos x = 1$ and $\cos(x\sqrt2) = 1$ hold true at the same time. The first statement is equivalent to $\frac{x}{\pi}\in\mathbb Z$, the second is equivalent to $\frac{x}{\pi}\sqrt2\in\mathbb Z$. The value $x=0$ fulfills the requirements. If a non-zero $x$ would fulfill both $\frac{x}{\pi}\in\mathbb Z$ and $\frac{x}{\pi}\sqrt2\in\mathbb Z$, the you could devide the second integer by the first. The result would be a rational number with value $\sqrt2$. This is a contradiction. Hence, only $x=0$ fulfills both requirements.