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This is a general question about using the $\mathcal{O}$ idea to say something about the sign of coefficient. Suppose I have $$x:=\frac{1}{b^2}\gamma(b)-\frac{1}{b^3}\beta(b)$$ and $b\in (b_0,+\infty)$ where $b_0 >1$. $\gamma(b)$ and $\beta(b)$ are bounded continous functions . $\gamma(b)$ bounded above by $\bar{\gamma}>0$ and below by $\underline{\gamma}>0$. $\beta(b)$ is also bounded above by $\bar{\beta}>0$ and bounded below by $\underline{\beta}<0$ . Can i say that the sign of $x$ will be positive for large $b$ . Or more precisely, there exists a $\hat{b}$ such that for $b>\hat{b}$, the statement should hold?

My attempt: Since $\underline{\gamma}>0$, there exists a $\hat{b}$, such that $$\underline{\gamma}-\frac{\bar{\beta}}{\hat{b}}\ge0$$. Now for all $b>\hat{b}$, the expression will be positive. Is my reasoning okay?

kangkan Dc
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You can't say anything about the sign of $x$. Consider $x=\frac{\sin(b)}{b^2}$ and $\gamma(b)=\frac{1}{b^2}$.

Maxence1402
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  • clarification: is beta=sin(b) and gamma=1/b^2. And they are additive. Won the first term be much larger in magnitude than the second term for large b. ? – kangkan Dc Feb 22 '20 at 14:52
  • I was thinking of $\beta(b)=0$, as $\sin(b)=\mathcal{O}(1)$. – Maxence1402 Feb 23 '20 at 10:06
  • I provided more information of the functions involved. – kangkan Dc Feb 25 '20 at 12:38
  • With the correct statements, you found the answer. Notice that you didn't use the $\mathcal{O}$. Be careful with them. They're useful for calculations, but not so much for properties. – Maxence1402 Feb 26 '20 at 21:56