why is f(n) equal to O(g(n)).. they could have made up a new symbol to establish such a relationship...
It is an equality relationship why would they need a new symbol?
f(n) = O(g(n)) .. n^2 = O(n^2) and n = O(n^2), so n = n^2 which is not making any sense
You are right it doesn't make any sense. Try n^2 = O(Selection_sort(n)),O(Bubble_sort(n)) = O(Selection_sort(n)), n^2 = O(Bubble_sort(n)). Nice and transitive the way it should be.
It's not an equality relationship. If it was an equality relationship it would be transitive. But as your parent points out, you have n=O(n^2), n^2=O(n^2) but not n=n^2, so transitivity doesn't hold.
Another way to see that the relationship can't be equality: the left hand side is a function, the right hand side is a set of functions. Two things from different classes can't be equal.
Unfortunately this answer is incorrect and is a common misunderstanding of big-O. Selection sort is typically stated to be O(n^2), which is true, but the way that O works is that it's a bound from above. It means "the set of functions that asympotically are guaranteed to not take more time than".
Selection sort is also O(n^3) and O(n!) and O(n^2 log log log log log n). The correct operator for saying that f(x) is O(n) is the set exists operator ∊. O(n^2) is a set of functions that is a strict subset of the set of functions that are in O(n^3).
Selection sort is only big-theta(n^2), which indicates both the bounds are tight (eg for Selection sort it is true that it's asymptotic behavior is both f(n) <= an^2 + m AND f(n) >= bn^2 + n.
It is an equality relationship why would they need a new symbol?
f(n) = O(g(n)) .. n^2 = O(n^2) and n = O(n^2), so n = n^2 which is not making any sense
You are right it doesn't make any sense. Try n^2 = O(Selection_sort(n)),O(Bubble_sort(n)) = O(Selection_sort(n)), n^2 = O(Bubble_sort(n)). Nice and transitive the way it should be.