asymptotic tight bound for quadratic functions

Question

In CLRS (Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein), for a function

f(n) = an² + bn

Answer 1

There's nothing special about those particular constants (other than the fact that in context of a certain n value they will satisfy the theta-ness of f). No magic. If you can find other positive constants whereby the relation holds that is just as valid (in fact, c1 can be ka for any 0). Although since they're there, let's analyse c1:

We need it to satisfy the following inequality:

c1*n^2 <= an^2 + bn + c

Let's take their value: c1 = a/4. For what n can we guarantee that the inequality holds? We could solve:

a/4*n^2 <= an^2 + bn + c
<==> 0 <= 3a/4*n^2 + bn + c

The quadratic has solutions at (-b +- sqrt(b^2-3ac)) / (3a/2), only the positive of which is of any interest since we have a positive leading coefficient polynomial, so we require n > 2 * (sqrt(b^2-3ac) - b) / 3a which is well defined assuming b^2 >= 3ac (and if not, then the quadratic is positive definite, which is even better, since its >=0 everywhere and the inequality holds for all n). I should point out that this is a valid solution, although we'll do a bit more work until we arrive at the book's representation.

We can split our analysis into 2 cases. First, let's assume |b|/a >= sqrt(|c|/a). So we can bound from above the inside of the sqrt(...) with 4b^2, and require n > 2/3 * [sqrt(4b^2)-b]/a which can be upper bounded by 2/3 * 3|b|/a = 2|b|/a.

Second case, let's assume |b|/a < sqrt(|c|/a). So we can bound from above the inside of the sqrt(...) with 4a|c|, and require n > 2/3 * [sqrt(4a|c|)-b]/a which can be upper bounded by 2/3 * 3*sqrt(a|c|)/a = 2sqrt(|c|/a).

So in each case we see that when we take max(|b|/a, sqrt(|c|/a)) our inequality holds when n > 2 * max

Similarly for c2.