Let's assume that we have a f(n) method, which we want to represent with a time complexity function (that is, a Set) g(n):
f(n) is O(g(n)) if, and only if, there exists constants c and n0, where f(n) <= cg(n) and where the input size n >= n0.
Now, let's try to apply this to our previous example:
f(n) = Tdouble = C5 * n + C4
f(n) = Tdouble = 4n + 1 // cause C5 and C4 can be any constants
For this example, we denoted it with the set O(n), that is, g(n) = n.
For our time complexity assertion to hold, we will need to satisfy the following condition:
4n + 1 <= c * n , where n >= n0
This equation is satisfied for values c = 5 and n0 = 1. Also, since the definition is met, we can safely say that the f(n) function is big-O(g(n)), that is, O(g(n)) or, in this case, O(n). We can also see this when plotted on a graph, as shown in the following diagram; after the value of n = 1, we can see that the value of c * g(n) is always greater than f(n) asymptotically. Take a look at the following diagram:
