What is the importance of c in big O notation?
Let's suppose the constant $c$ wasn't there: let's say that $f(n)=O'(g(n))$ if $f(n)\leq g(n)$ for all large enoun&nbsp;$n$. So, for example, $n=O'(n^2)$ and $\log n = O'(n)$, as you'd expect. But let's look at some functions whose growth rates are closer. First, let $f(n)=n+1$ and $g(n)=n$: well, $f(n)\neq O'(g(n))$ because $n+1 \not\leq n$ for all&nbsp;$n$.  Similarly, $2n\neq O'(n)$. SO, we see that the  constant&nbsp;$c$ is absolutely essential to the definition of&nbsp;$O(-)$: it just doesn't do what we want it to do if we remove it.

$O(-)$ measures the growth rate of functions ignoring constant factors. It gives you notation to say things like "$f$ is linear" or "$f$ is quadratic".  When we say "$f$ is linear", we don't distinguish between, say, $f(n)=n$ or $f(n)=4n$ or $f(n)=2n+12$: we ignore the constants. The constant&nbsp;$c$ in the definition of $O(-)$ allows it to ignore multiplicative constants and additive terms.