Is $\{\sin^n{(n)}:n\in\mathbb{N}\}$ dense in $[-1,1]$?

Solution 1:

It is very likely that one cannot prove anything rigorous, but here is at least a heuristic showing that the answer to your question is likely to be positive; that is, $\sin^n(n)$ is dense in $(-1,1)$.

We want to show that any fixed subinterval $(\alpha,\beta)\subseteq(-1,1)$ contains a number of the form $\sin^n(n)$. Assume for simplicity that $\alpha>0$. Then $\sin^n(n)\in(\alpha,\beta)$ means that $\sin(n)\in(\alpha^{1/n},\beta^{1/n})$. Using Taylor's approximation, the interval $(\alpha^{1/n},\beta^{1/n})$ has length $C(1+o(1))n^{-1}$, where $C=\log(\beta/\alpha)$. Assuming uniform "distribution" of $\sin(n)$ in $[-1,1]$, the "probability" that this happens is $0.5C(1+o(1))n^{-1}$. The sum of these "probabilities" diverges (as so does the harmonic series $\sum n^{-1}$), and this shows that the event in question is likely to happen infinitely often; that is, our interval $(\alpha,\beta)$ is likely to contain infinitely many numbers $\sin^n(n)$.