# Exponential Sinusoidal Function (with Damping) and the Distances of Zeros

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!

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 $$(-\infty, 0)$$
• Forgot to add - obviously include the $e^{-x}$ envelope to get the damping Oct 13, 2021 at 18:53