Sum of cosines with different $\omega$

Question

Let $S$ be a sum of cosines with the same amplitude and different $\omega$:

$$S(x) = A_0 \sum_{n = 1}^N \cos \left( \omega_n x + \phi_n \right)$$

that is $\omega_1 \neq \omega_2 \neq \ldots \neq \omega_n$.

What are the conditions to be verified for the $\omega_n$ and the $\phi_n$ in order to obtain, for at least one value of $x$, that $S(x) = NA_0$?

That is: when is it true that, given a number $N$ of summed cosines of amplitude $A_0$, the result will reach (at least once) the value $NA_0$?

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For $N = 2$, a sufficient condition to be satisfied is: $\omega_1 = \omega_0 + \alpha$ and $\omega_2 = \omega_0 - \alpha$, along with $\phi_1 = \phi_0 + \beta$ and $\phi_2 = \phi_0 - \beta$. In that case, $S(x) = 2A_0 \cos \left( \omega_0 x - \phi_0 \right) \cos \left( \alpha x - \beta \right)$.

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I would be pleased if who downvoted this question will leave a comment with a suggestion.

Answer 1

If you are free to adjust the phases, then set $\phi_n=0$ for all $n$ and trivially

$$S(0)=NA_0.$$

If not, you can locate the extrema by canceling the first derivative. This gives the difficult equation

$$\sum_{n=0}^N\omega_n\sin(\omega_n x+\phi_n)=0.$$

This equation has infinitely many solutions which appear in a chaotic way when the $\omega$ are irrational mutiples of each other. I suspect that the sum never reaches $NA_0$ exactly (@Ingix proved it when $N=2$), but gets as close as you want for larger and larger $x$. But I can't prove this stronger property.

A case with $N=2$:

Don't believe that $2$ is ever exactly reached.

Answer 2

Your proposition isn't even true for $N=2$. Set $\omega_1=1, \omega_2=2, \phi_1=\phi_2=\sqrt{2}$. In order for each cosine to be one, the argument must be an integer multiple of $2\pi$:

$$\begin{array}{c} \omega_1 x + \phi_1=2k_1\pi \\ \omega_2 x + \phi_2=2k_2\pi \end{array}$$

Subtracting the first equation from the second, one gets $ x = 2(k_2-k_1)\pi$ and putting the back into the first equation

$$ \sqrt{2}=\phi_1=2k_1\pi-2(k_2-k_1)\pi=(4k_1-2k_2)\pi$$

which would imply $\sqrt{2}$ to be an integer multiple of $\pi$, which is of course impossible.