Writing a proof of phase difference between 2 sine waves - necessary or self evident?

Question

thanks for reading. edited based on helpful comments

Using GeoGebra to add sine waves with the same amplitude together to produce a graph of superposition is able to generate a pseudo-random pattern

I would now like to express this more formally.

I would like to prove that given an infinite set of sine waves, with different frequencies, but with the same amplitude, when summed to create a function, will have a rate of change that always produces minimums.

Reason being I want to prove that the set of pseudo-randomly occurring minimums is also infinite.

The sum of the functions is this;

$ f(x) = \sum_1^\infty [sin^2( \frac x2 \pi) + sin^2( \frac x{2n+1} \pi)] $

and hence the rate of change is $ f'(x) $

Answer 1

Here's a statement that I think comes close to what you're trying to say about "constructive interference".

Claim: Let $\alpha,\beta \neq 0$ be such that $\alpha/\beta$ is irrational. Let $f:\Bbb R \to \Bbb R$ be given by $f(x) = \sin(\alpha x) + \sin(\beta x)$. Let $M$ denote the supremum $M = \sup_{x \in \Bbb R} f(x)$. For any $\epsilon > 0$, let $S_\epsilon$ denote the set of $x$ for which $f(x) > M- \epsilon$.

$M = 2$, and for any $\epsilon > 0$, the set $S_\epsilon$ is infinite and unbounded.

Sketch of proof: First, we note that $M \leq 2$, since $$ M = \sup_{x \in \Bbb R}(\sin \alpha x + \sin \beta x) \leq \sup_{x \in \Bbb R}\sin\alpha x + \sup_{x \in \Bbb R} \sin \beta x \leq 1 + 1 = 2. $$

We note that $\sin(\alpha x)$ reaches its maximum value at $x_n = \frac{(2n + 1) \pi}{2 \alpha}$ for integer values of $n$. It suffices to show that for any $0 < \epsilon \leq 1$, there exist infinitely many $n \in \Bbb Z$ such that $\sin(\beta x_n) > 1 - \epsilon$. Let $\langle x\rangle$ denote the integer-part of $x$, i.e. $\langle x\rangle = x - \lfloor x \rfloor$. It can be shown that the sequence $\langle \frac{x_n}{2 \pi \beta} \rangle $ is dense in $[0,1]$ (see this post and this post, for instance). From there, it follows that there are infinitely many $n$ for which $$ \frac{\arccos(1 - \epsilon)}{2 \pi} < \frac{x_n}{2 \pi \beta} < \pi - \frac{\arccos(1 - \epsilon)}{2 \pi}, $$ and the conclusion follows.

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We could show that $f$ also comes arbtirarily close to its infimum by noting that $f(-x) = -f(x)$ and applying the above result.