Question

Suppose f(n) = O(s(n)), and g(n) = O(r(n)). All four functions are positive-valued and monotonically increasing. Prove (using the formal definitions of asymptotic notations) or disprove (by counterexample) each of the following claims:

(a) f(n) − g(n) = O(s(n) − r(n))

(b) if s(n) = O(g(n)), then f(n) = O(r(n))

(c) if r(n) = O(s(n)), then g(n) = O(f(n))

(d) if s(n) + g(n) = O(f(n)), then f(n) = Θ(s(n))

Answer 1

Required defintion -

a(n) = O(b(n)) if there exists a positive integer n0 and a positive constant c, such that a(n)≤c.b(n) ∀ n≥n0

=> for f(n) = O(s(n)), f(n) <= c1.s(n) , n >= n10 - eqn 1

and for g(n) = O(r(n)), g(n) <= c2.r(n) , n >= n20   - eqn 1

Also, to setlle on common n0 value, take n0 = max(n10, n20) and for c, c = max(c1, c2)

(a) f(n) − g(n) = O(s(n) − r(n))

Subtracting eqn 2 from eqn 1,

f(n) - g(n) <= c(s(n) - r(n)), n >= n0

=> this can be written as - f(n) − g(n) = O(s(n) − r(n))

Hence proved.

(b) if s(n) = O(g(n)), then f(n) = O(r(n))

s(n) = O(g(n))

=> s(n) <= c3.g(n) , n >= n0

From eqn 1 and 2,

f(n) <= c1.s(n) <= c1.c3.g(n) <= c1.c3.c2.r(n)

=> f(n) <= c1.c3.c2.r(n)

=> f(n) <= c.r(n) for c = c1.c3.c2

=> f(n) = O(r(n))

Hence proved.

(c) if r(n) = O(s(n)), then g(n) = O(f(n))

r(n) = O(s(n))

=> r(n) <= c3.s(n) , n >= n0

g(n) <= c2.r(n) <= c2.c3.s(n)

The function f(n) does not fit in the equation => Disprove the statement.

Example -

Let f(n) = 1

s(n) =

g(n) = n

r(n) =

Thus, all the conditions fit, that is, f(n) = O(s(n)), g(n) = O(r(n)) and r(n) = O(s(n))

But g(n) O(f(n)), instead f(n) = O(g(n))

=> g(n) = O(f(n)) is not always true.

(d) if s(n) + g(n) = O(f(n)), then f(n) = Θ(s(n))

s(n) + g(n) = O(f(n))

=> s(n) + g(n) <= c3.f(n)

And we know that, f(n) <= c1.s(n)

Adding the function g(n), we only increase (or remain constant) the growth of the function s(n).

=> s(n) + g(n) = O(f(n)) may not be true.

But if f(n) = Θ(s(n)), s(n) <= c4.f(n)

Then this can be true.

Hence proved.