Considering that $1\over{2\pi}$ is the frequency of $\cos x$ and also of $\cos x - 2\cos 2x$, what is the frequency of $\cos x - \sqrt 2\cos\sqrt 2 x$?


Perhaps "frequency" isn't the interesting quantity. I think the period is infinite. That would make the frequency zero. Maybe what I want to know is bandwidth?

I think the additive combination (superposition) of the two waves produces a component at the higher frequency $1/(2\pi(\sqrt 2-1))$. I would hope that sampling at a frequency of $1/(\sqrt 2\pi -\pi)$ would be sufficient to reveal as many zero-crossings as we may wish. Is that correct? Is it the best answer?

I understood Fourier series and partial differential equations when I took that course and I still get the general idea, but it has been a long time.


Mistakenly, I devised and posed this question for my colleagues as a weekly math challenge before I realized that I don't know how to get the answer myself.

I'll provide additionally just some background about my motive for asking this particular question. Maybe it can help you give an answer that is relevant to me. However, this background information in itself is not part of my question. So, I came up with the question about frequency hoping to move us nearer to another question that Mark Kac used in order to elucidate stochastic independence. That is, given any particular value $-2 <\alpha < 2$, in what fraction of the interval $0\leq x < 100$ (just to choose some arbitrary bounds) is $f(x) <\alpha$ where $f(x) = \sin x - \sin\sqrt 2 x$? The interesting destination that I want to reach is that the answer to this further question is (somehow related to) the normal distribution function, yet there is nothing "random" involved.

Of course $f'(x) =\cos x -\sqrt 2\cos\sqrt 2 x$. If I know the "frequency" (or something) of $f'(x)$ then I hope I can find, at least numerically, its zero-crossings. Those include all of the local extrema of $f(x)$. Then, knowing that the sum-to-product formula gives $f(x) = 2\cos(x/2+x/\sqrt 2)\sin(x/2-x/\sqrt 2)$, I think that my favorite computer algebra system and I can find, at least numerically, where $f(x) <\alpha$.


1 Answer 1


Not sure, whether it's really your question. The signal for which you asked for the frequency actually has two frequencies. There are two components, the two cosines. And each component has a frequency. Assume your frequency is called $f$. So, the signal looks like $\cos(2\pi f x)$. So, getting the frequency of a component is quite easy. Just take everything from it between the brackets and divide it by $2\pi$.

Probably, you already got the frequencies of your signal. But just to be sure, it's: $\frac{1}{2\pi}$ and $\frac{\sqrt{2}}{2\pi}$.

Hope, it was the answer, you're looking for...

  • $\begingroup$ +1 but I think the definition of frequency should be $(\mathrm d\phi/\mathrm d x)/(2\pi)$ where $\phi$ is the argument of the cosine. $\endgroup$
    – Deve
    Nov 1, 2013 at 15:09
  • $\begingroup$ Yes. Wanted to keep it as simple as possible. But, you're right, my answer is only correct for linear arguments in the sine/cosine. @ Peter K.: Thanks for making the formulas looking nice! $\endgroup$
    – Pascal
    Nov 1, 2013 at 15:26

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