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So I tried to solve it but got confused because I got weird ressult.

$n=O(n+n\sin(n))$

$n≤c(n+n\sin(n)) \quad c=1$

$n≤n+n\sin(n)$

$1≤1+1\sin(n)$

$0≤\sin(n)$

I am wondering am I right to prove that equation or I should disprove it and how to deal with disproving?

ACB
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BMW BOI
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    Hint 1: This is not true, try to disprove it. Hint 2: Whenever $\sin n\approx -1$, the value $n+n\sin n$ is much smaller than $n$. –  Jan 14 '22 at 13:16
  • where does this problem come from? According to Wiki it is requested that the function in th O-term is "strictly positive for all large enough values of x". This is not the case in your example. – miracle173 Jan 14 '22 at 13:19
  • @miracle173 $n+n\sin(n)=n(1+\sin(n))>0$ because $\sin(n)=-1$ only for $n=\pi(2k-\frac{1}{2}), k\in\mathbb Z$, which is irrational (and certainly not an integer) for all $k$. (I assume in this question $n$ is an integer.) –  Jan 14 '22 at 13:59

1 Answers1

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You want to find an $C>0$ and $N>0$ such that

$$n\le C|n+n\sin(n)|, \forall n>N \tag 1$$ From this equation we get

$$n\le Cn|1+\sin(n)|$$ We can cancel $n$ and because $\sin(x)\ge -1, \forall x \in \mathbb R$ we get $$1\le C(1+\sin(n))$$ and further $$-(1-\frac 1 C) \le \sin(n)$$

But the Equidistribution theorem says that

$$\frac{n}{2\pi} \pmod 1, n\in \mathbb N$$ is uniformly distributed in $(0,1)$. Multiplyiong these numbers bei $2\pi$ shows that $$n \pmod {2\pi}$$ is uniformly distributed in $(0,2\pi)$ and so for each $C$ there are $n$ such that $$-1<\sin(n)<-1+\frac 1 C$$ So $(1)$ does not hold for any $C$.

Annotation:

A similar result can be derived from Dirichlet's approximation theorem

miracle173
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