# What is the importance of c in big O notation?

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?

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.