When we solve the recursive functions using substitution method, the impact of ceil and floor functions is trivial when the size of the input is large enough. For example the answer of$$T(n) = T(\lceil n/3 \rceil) + O(n)$$is equivalent to$$T(n) = T(n/3) + O(n)$$when the $n$ is large enough. Since $$T(n) = T(\lceil n/3 \rceil) + O(n) \leq T((n/3) + 1) + O(n)$$and the size $1$ added to $n/3$ doesn't effect the running time of the algorithm. My question is, when the ceil and floor functions don't effect the asymptotic bounds of a algorithm, why not to ignore them and simply solve the $T(n)$ without them?CLRS page 67 edition3 saysWhen we state and solve recurrences, we often omit floors, ceilings, and boundary  conditions.
Why do we count the ceils and floors in recursive functions?
algorithm analysis;asymptotics;recurrence relation;recursion