Is this a valid use of big-O notation? Suppose that $m=O(n^{c+1/2})$ for some real $c&gt;0$ and $x=O(\sqrt{\log m})$. Are the following two computations valid? I understand that I'm abusing notations a bit to get at the desired results.

Computation 1
\begin{align}
f(x)\leq \exp(-x^2/2)&amp;=\exp(-O(\log m))\\
&amp;=\exp(-O(\log O(n^{c+1/2})))\\
&amp;=\exp(-O(\log n^{c+1/2}))\\
&amp;\leq \exp(-K_1\log n^{c+1/2})))\\
&amp;=n^{-O(1)}
\end{align}
where $K_1&gt;0$ when $n$ is sufficiently large (i.e. there exists $K_1&gt;0$, $n_0$ such that when $n\geq n_0$, $\exp(-O(\log n^{c+1/2}))\leq  \exp(-K_1\log n^{c+1/2})))$. The last quantity $n^{-O(1)}$ represents the set of polynomially-bounded functions (see accepted answer in this post).

Computation 2
$$
f(x)\frac{m}{n^{1/2}} \leq n^{-O(1)}\frac{m}{n^{1/2}}=\leq n^{-O(1)}\frac{m}{n^{c+1/2-c}}=O(1)??
$$
If these computations appear in a paper for me to review, I would like to ask the author to rewrite.


  $$\exp(-x^2/2)\leq \exp\left(-K_1\log n^{c+1/2\,}\right), \text{ where } K_1&gt;0$$


This is wrong. The direction of inequality should be switched. For example, let $n\to\infty$ and $x$ be constant 1, the left hand side remains a constant while the right hand side goes to 0. This error means "Computation 2" is wrong from the start.


  The last quantity $n^{-O(1)}$ represents the set of polynomially-bounded functions.


Instead of $n^{-O(1)}\,$, $n^{O(1)}$ represents asymptotically the set of polynomially-bounded functions.