Hey Sneaky.
Can there exist a c2 such that c2 > 1?
Maybe you should outline what processes have the property where c1 and c2 both equal 1 and then say that the complement of processes of this set of processes don't follow that.
If g(n) is a upper bound of f(n), then g(n) is not a lower bound on f(n).
I think this statement above is false because if f(n) ∈ theta(1), which means the running time of f(n) is constant (ex. f(n) = 1), then this is a counter example of the above statement.
If c_{1}(g(n)) <= f(n) <= c_{2}(g(n))
then let g(n), c_{1}, c_{2} = 1.
1<=1<=1
which is true for all n >= x_{0} for some x_{0} > 0.
Is this correct?
Hey Sneaky.
Can there exist a c2 such that c2 > 1?
Maybe you should outline what processes have the property where c1 and c2 both equal 1 and then say that the complement of processes of this set of processes don't follow that.
So for example, if f(n) = n, which is a linear equation, then the upper bound can equal the lower bound right? But I don't understand if there is a function which the upper bound has to equal the lower bound at all times?