Eliminating order notation in upper bound

Question

I have that some value $E_i=\alpha^2\varepsilon_i^3+O(\varepsilon_i^4)$, where $\alpha>0$ is a fixed constant and for every $i$, $0<\varepsilon_i\ll1$. I would like to place an upper bound on $E_i$ that does not include the term $O(\varepsilon_i^4)$ by multiplying $\alpha^2\varepsilon_i^3$ with some constant. The result I end up with is $E_i\leq2\alpha^2\varepsilon_i^3$ through the following reasoning. \begin{eqnarray} c\cdot\alpha^2\varepsilon_i^3&\geq&\alpha^2\varepsilon_i^3+O(\varepsilon_i^4)\\ (c-1)\alpha^2\varepsilon_i^3&\geq&O(\varepsilon_i^4)\\ c-1&\geq&\frac{O(\varepsilon_i^4)}{\alpha^2\varepsilon_i^3}=O(\varepsilon_i)\\ c&\geq&1+O(\varepsilon_i). \end{eqnarray} Is this reasoning correct? $c$ does not have to be optimal but is there a way to get a smaller constant?

Answer 1

The reasoning is incorrect for two different reasons.

Firstly if you want to make reversible deductions you have to check and state so. I doubt you checked so it is probably by luck that your steps are all reversible.

Secondly after you got to the last line $c \ge 1 + O(ε)$ (I shall drop the unused subscripts) you made an incorrect deduction to get $c \ge 2$. It is true that the latter implies the former but not the other way around. In fact the former is equivalent to $c > 1$ (again you should check both directions). To prove the reverse direction check that $c-1 > 0$ implies $c-1 \gg ε$ by definition of $\gg$. Note that the forward direction fails if you only have $0 \le ε \ll 1$ since it would then be possible for $ε$ to be zero and so all we can get is $c \ge 1$.