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Could anyone tell me what function could this be? Basically it's a sinusoidal function, with some damping, which I believe have component of trigonometric $\sin(x)$ or $\cos(x)$ and the exponential $e^{-x}$. And the zeros of this function would be way farther apart as $x \rightarrow -\infty$ than if it tends $x \rightarrow \infty$. But I think in the actuality because of the exponential decay, then $x \rightarrow 0$ as $x \rightarrow \infty$ so the distances of the zeros of the function seem to approach $0$ as $x \rightarrow \infty$. I'm thinking it could be like $$f(x) = e^{-x}\cos(2\pi x)$$ but by looking at the graph it seems like that the distances of the zeros of the function are rather the same when $x \rightarrow -\infty$ and what I want is that the distances of the zeros of the function grow larger and larger as $x \rightarrow -\infty$.

Anyone knows how I can construct this function? Thanks so much!

P.S. I'm not sure if I was explicit with what I mean here by "distances of zeros". So for example $f(x) = (x-1)(x-2)(x-5)$ has three zeros: $1, 2, 5$. The zero $1$ and the zero $2$ have distance $|1-2| = 1$ while the zero $2$ and $5$ have distance of $|2-5| = 3$. For convenience, we only consider the distances of consecutive zeros; so in this example, the distance of $1$ and $2$ are considered, as well as of $2$ and $5$, but we do not make anything out of the distance of $1$ and $5$ because they are not consecutive. I probably have not made myself clear here due to the informality of the wordings, but let me know if there is something in the question that is unclear thanks!

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Yeah, you can use $cos(2\pi\sqrt{x})$ if you want the zeroes to get further apart as you go to infinity. You can also use the cubed root if you need the function defined on $(-\infty, 0)$

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  • $\begingroup$ Forgot to add - obviously include the $e^{-x}$ envelope to get the damping $\endgroup$
    – WhoDatBoy
    Oct 13, 2021 at 18:53

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