Prerequisite: Asymptotic Notations
Assuming f(n), g(n) and h(n) be asymptotic functions the mathematical definitions are:

1.      If f(n) = Θ(g(n)), then there exists positive constants c1, c2, n0 such that 0 ≤ c1.g(n) ≤ f(n) ≤ c2.g(n), for all n ≥ n0

2.      If f(n) = O(g(n)), then there exists positive constants c, n0 such that 0 ≤ f(n) ≤ c.g(n), for all n ≥ n0

3.      If f(n) = Ω(g(n)), then there exists positive constants c, n0 such that 0 ≤ c.g(n) ≤ f(n), for all n ≥ n0

4.      If f(n) = o(g(n)), then there exists positive constants c, n0 such that 0 ≤ f(n) < c.g(n), for all n ≥ n0

5.      If f(n) = ω(g(n)), then there exists positive constants c, n0 such that 0 ≤ c.g(n) < f(n), for all n ≥ n0

Properties:

1.      Reflexivity:
If f(n) is given then

f(n) = O(f(n))

Example:
If f(n) = n³ ⇒ O(n³)
Similarly,

f(n) = Ω(f(n))

f(n) = Θ(f(n))

2.      Symmetry:

f(n) = Θ(g(n)) if and only if g(n) = Θ(f(n))

Example:
If f(n) = n² and g(n) = n² then f(n) = Θ(n²) and g(n) = Θ(n²)
Proof:

·         Necessary part:
f(n) = Θ(g(n)) ⇒ g(n) = Θ(f(n))
By the definition of Θ, there exists positive constants c1, c2, no such that c1.g(n) ≤ f(n) ≤ c2.g(n) for all n ≥ no
⇒ g(n) ≤ (1/c1).f(n) and g(n) ≥ (1/c2).f(n)
⇒ (1/c2).f(n) ≤ g(n) ≤ (1/c1).f(n)
Since c1 and c2 are positive constants, 1/c1 and 1/c2 are well defined. Therefore, by the definition of Θ, g(n) = Θ(f(n))

·         Sufficiency part:
g(n) = Θ(f(n)) ⇒ f(n) = Θ(g(n))
By the definition of Θ, there exists positive constants c1, c2, no such that c1.f(n) ≤ g(n) ≤ c2.f(n) for all n ≥ no
⇒ f(n) ≤ (1/c1).g(n) and f(n) ≥ (1/c2).g(n)
⇒ (1/c2).g(n) ≤ f(n) ≤ (1/c1).g(n)
By the definition of Theta(Θ), f(n) = Θ(g(n))

3.      Transistivity:

f(n) = O(g(n)) and g(n) = O(h(n)) ⇒ f(n) = O(h(n))

Example:
If f(n) = n, g(n) = n² and h(n) = n³
⇒ n is O(n²) and n² is O(n³) then n is O(n³)
Proof:
f(n) = O(g(n)) and g(n) = O(h(n)) ⇒ f(n) = O(h(n))
By the definition of Big-Oh(O), there exists positive constants c, no such that f(n) ≤ c.g(n) for all n ≥ no
⇒ f(n) ≤ c1.g(n)
⇒ g(n) ≤ c2.h(n)
⇒ f(n) ≤ c1.c2h(n)
⇒ f(n) ≤ c.h(n), where, c = c1.c2 By the definition, f(n) = O(h(n))
Similarly,

f(n) = Θ(g(n)) and g(n) = Θ(h(n)) ⇒ f(n) = Θ(h(n))

f(n) = Ω(g(n)) and g(n) = Ω(h(n)) ⇒ f(n) = Ω(h(n))

f(n) = o(g(n)) and g(n) = o(h(n)) ⇒ f(n) = o(h(n))

f(n) = ω(g(n)) and g(n) = ω(h(n)) ⇒ f(n) = ω(h(n))

4.      Transpose Symmetry:

f(n) = O(g(n)) if and only if g(n) = Ω(f(n))

Example:
If f(n) = n and g(n) = n² then n is O(n²) and n² is Ω(n)
Proof:

·         Necessary part:
f(n) = O(g(n)) ⇒ g(n) = Ω(f(n))
By the definition of Big-Oh (O) ⇒ f(n) ≤ c.g(n) for some positive constant c ⇒ g(n) ≥ (1/c).f(n)
By the definition of Omega (Ω), g(n) = Ω(f(n))

·         Sufficiency part:
g(n) = Ω(f(n)) ⇒ f(n) = O(g(n))
By the definition of Omega (Ω), for some positive constant c ⇒ g(n) ≥ c.f(n) ⇒ f(n) ≤ (1/c).g(n)
By the definition of Big-Oh(O), f(n) = O(g(n))

Similarly,

f(n) = o(g(n)) if and only if g(n) = ω(f(n))

5.      Since these properties hold for asymptotic notations, analogies can be drawn between functions f(n) and g(n) and two real numbers a and b.

·         g(n) = O(f(n)) is similar to a ≤ b

·         g(n) = Ω(f(n)) is similar to a ≥ b

·         g(n) = Θ(f(n)) is similar to a = b

·         g(n) = o(f(n)) is similar to a < b

·         g(n) = ω(f(n)) is similar to a > b

6.      Observations:

max(f(n), g(n)) = Θ(f(n) + g(n))

Proof:
Without loss of generality, assume f(n) ≤ g(n), ⇒ max(f(n), g(n)) = g(n)
Consider, g(n) ≤ max(f(n), g(n)) ≤ g(n)
⇒ g(n) ≤ max(f(n), g(n)) ≤ f(n) + g(n)
⇒ g(n)/2 + g(n)/2 ≤ max(f(n), g(n)) ≤ f(n) + g(n)
From what we assumed, we can write
⇒ f(n)/2 + g(n)/2 ≤ max(f(n), g(n)) ≤ f(n) + g(n)
⇒ (f(n) + g(n))/2 ≤ max(f(n), g(n)) ≤ f(n) + g(n)
By the definition of Θ, max(f(n), g(n)) = Θ(f(n) + g(n))

7.      O(f(n)) + O(g(n)) = O(max(f(n), g(n)))

Proof:
Without loss of generality, assume f(n) ≤ g(n)
⇒ O(f(n)) + O(g(n)) = c1.f(n) + c2.g(n)
From what we assumed, we can write
O(f(n)) + O(g(n)) ≤ c1.g(n) + c2.g(n)
≤ (c1 + c2) g(n)
≤ c.g(n)
≤ c.max(f(n), g(n))
By the definition of Big-Oh(O),
O(f(n)) + O(g(n)) = O(max(f(n), g(n)))