What is the importance of c in big O notation?

Question

From the definition of Big O, it states that there should be a function $g(x)$ such that it is always greater than or equal to $f(x)$. Or $f(x) \le cg(n)$ for all values of $n > n_0$. What I'm not able to understand is what is its importance in Big oh notation. Every question explains what is big-Oh and what how it uses useful. But I'm not able to understand what is $c$ actually in big-oh. There might be answer which I'm not able to get it due to its technical jargon.

Can someone explain me this in simple terms?

Answer 1

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 $n$. 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 $n$. Similarly, $2n\neq O'(n)$. SO, we see that the constant $c$ is absolutely essential to the definition of $O(-)$: 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 $c$ in the definition of $O(-)$ allows it to ignore multiplicative constants and additive terms.