Define: Let $f,g:\mathbb N\to \mathbb R^+$, $g(n)=\hat O f(n)$ if $\exists c>0: \forall n\ge 1 : g(n)\le cf(n)$.
Prove that $\hat O$ and the definition of big $O$ are equivalent, i.e: $g(n)=\hat O f(n) \iff g(n)= O f(n)$
One side is simple, just take $n_0=1$, but how do I show that $g(n)=\hat O f(n) \Leftarrow g(n)= O f(n)$?
Suppose $\exists n_0,c>0: \forall n \ge n_0 :g(n)\le c f(n)$, we can't know what happens between $1$ and $n_0$. I thought about moving the functions by defining $x=n-n_0$ and then plugging $x$ to the functions, now we will have the requirements of $\hat O$ but it feels like cheating.
PS: I'm not sure what tags this should have.
You have the constant that works for $n_0$ and you know that below $n_0$ you just have two values $g(n)$ and $f(n)$ so you can use their quotient to give $c_n$. So you have $c_i$ for $1...n_0$ and the c from the big O definition. Cook up a new c from that finite set.