How to Prove $n=O(100n+n\sin(n))$

Asked Jan 12 '22 at 21:42

Active Jan 12 '22 at 21:45

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So I started learning about O notation and got this question

Prove $n=O(100n+n\sin(n))$.

I found it hard but still tried my best.

\begin{align} n&=O(100 n+n \sin(n)),\\ n &\le c( 100n+n\sin(n)) , \ \ c=1 \\ n &\le 100n+n\sin(n) \\ 1 &\le 100+\sin(n) \\ -99 &\le \sin(n), \ \ N=1. \end{align} like I know it looks like piece of cake but $\sin(n)$ made me wonder if it's a right way to solve and prove $O$ notation including $\sin(n)$?

edited Jan 12 '22 at 21:45

Arctic Char

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asked Jan 12 '22 at 21:42

BMW BOI

2

From $-1\leq \sin n\leq 1,$ deduce:$$99\leq 100+\sin n\leq 101.$$ Multiply by $n$ and note $n<99n.$ – Thomas Andrews Jan 12 '22 at 21:51
For $n\geq 1$, $n < 99n = 100n - n \le 100n + n\sin n$, since $-1 \le \sin n\le 1$. – Gary Jan 13 '22 at 01:38

How to Prove $n=O(100n+n\sin(n))$

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