Big O: find ##c## and ##n_0## if ##g(n)## is known

[Korisnik]

TL;DR

easy way to find c and n0 if g(n) known in O(g(n))

Is it guaranteed that when I have any kind of function $f$, that I'll be able to find constants $c$ and $n_0$ if I know $g$ by simply adding up the factors multiplying all the terms in the equation?

Suppose $f(n) = a\cdot n^{143} + b \cdot n! + d \cdot (n!)! + m \cdot 2^n - p \cdot \sqrt{n} + 42$, now I know here that $f$ is $\mathcal{O}((n!)!)$, meaning that $g(n) = (n!)!$ (where $\mathcal{O}$ takes $g(n)$).

But can I immediately assume that that holds if I pick $c = a + b + d + m + p + 42$ (taking absolute values of all the factors multiplying each term in $f(n)$) and letting $n_0 = 1$ such that $0 \leq f(n) \leq c \cdot g(n), \ \forall n\in\mathbb{Z}^+,\ n\geq n_0$?

Will this technique work for all functions $f$? Moreover, I think this can be simplified further by not even taking the factors before negative terms (in this case $p$).

Thanks in advance.

Edit:

Actually, I think this is sort of trivial to prove, because if indeed we have the "biggest" function $g(n)$ figured out, then if we simply do the following: $$0 \leq f(n) = a\cdot n^{143} + b \cdot n! + d \cdot (n!)! + m \cdot 2^n - p \cdot \sqrt{n} + 42 \leq c \cdot g(n) = (a + b + d + m + p + 42) \cdot (n!)! = a \cdot (n!)! + b \cdot (n!)! + d \cdot (n!)! + m \cdot (n!)! + p \cdot (n!)! + 42 \cdot (n!)!$$ and each of these new terms grows faster than each of the terms in $f(n)$, and further, since negative terms only "discontribute", there's no need to even include those, because by excluding them, you get a faster growing function than $f(n)$.

 

[mfb]

With n₀=1 this won't work.

Consider f(x)=x!+2^x. Your approach would lead to g(x)=2x!, then g(1)=2 but f(1)=3.

The approach works if you choose an n₀ such that the other terms have the asymptotic inequalities, in this case $x! \geq 2^x$.