Homework 1 Solutions
50 points
Define a function f to be admissable iff it is defined on almost all of the nonnegative integers and has non-negative real values on almost all of the domain.
That is, there is an integer n0 such that f (n) is defined and f (n) ≥ 0 for all
integers n > n0 . Note that for polynomials, this condition requires that the
leading coefficient is a positive number.
Proposition 1. If f and g are admissable and f ≤ O(g) then Θ(f + g) = Θ(g).
Proof. Applying the definition of big-O, we find that there is a positive constant
C and a positive integer n1 such that
f (n) ≤ Cg(n)
for n ≥ n1 . Therefore we have
f (n) + g(n) ≤ Cg(n) + g(n) = (C + 1)g(n)
for n ≥ n1 . Taking C1 = 1 + C we have
f (n) + g(n) ≤ C1 g(n)
and thus f + g ≤ O(g).
On the other hand, note that by admissibility there exists a positive integer n2
such that f (n) ≥ 0 and therefore
g(n) ≤ f (n) + g(n)
for all n ≥ n2 . Taking C2 = 1, we then have
C2 g(n) ≤ f (n) + g(n)
for all n ≥ n2 and thus f + g ≥ Ω(g). It now follows that f + g = Θ(g).
Exercise. Prove or disprove: Θ(1 + g) = Θ(g) for any admissable g.
Solution. This statement is false. Reading Prop 1 above, we see that a counterexample must satisfy g = o(1), that is, g(n) → 0 as n → ∞. Any such g will do - for
example, g(n) = 1/n. There is no C > 0 such that g(n) ≥ C for almost all n.
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Proposition 2. Suppose that f and g are admissable and
f (n)
lim
=C
n→∞ g(n)
where C is a non-negative constant. Then f = O(g) and g = Ω(f ). Moreover, if
C > 0 then f = Θ(g).
Proof. By the definition of limit, f (n)/g(n) → C, with = 1, there exists a
positive integer n1 such that f (n)/g(n) ≤ C + 1 for n ≥ n1 . Taking C1 = 1 + C we
have f (n)/g(n) ≤ C1 and after algebra
f (n) ≤ C1 g(n)
for n ≥ n1 . Therefore f ≤ O(g).
If in addition C > 0, again applying the definition of limit with = C/2, there is
a positive integer n2 such that f (n)/g(n) ≥ C/2 for n ≥ n2 . Taking C2 = C/2 we
have f (n)/g(n) ≥ C2 and after algebra
C2 g(n) ≤ f (n)
for n ≥ n2 . Therefore f ≥ Ω(g).
Proposition 3. If a and b are positive numbers (not necessarily integers) and a ≤ b
then na ≤ O(nb ) and nb ≥ Ω(na ).
Proof. Let f (n) = na and g(n) = nb , and consider the quotient:
f (n)
na
= b = na−b .
g(n)
n
Note that c = a − b ≤ 0 implies that nc → 0 as n → ∞. Applying Prop 2 above, we
conclude that f ≤ O(g).
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Proposition 4. Suppose p is a polynomial in n of degree d, with the form
p(n) =
d
X
ai ni
i=0
with leading coefficient ad > 0. Then p(n) = Θ(nd ).
Proof 1. Note that p(n) = q(n) + ad nd where
q(n) =
d−1
X
ai ni
i=0
If ad−1 < 0 then q is not admissible and moreover q(n) < 0 for almost all n. Otherwise
q ≤ O(ad nd ) by Prop 3. Therefore, by Prop 1 with g(n) = ad nd , we have
p = Θ(ad nd ) = Θ(nd ). Proof 2. Consider the quotient
d
p(n) X
ai ni−d
=
d
n
i=0
Note that all of the terms on the right have negative exponent except for the last one
ad nd−d = ad which is a positive constant. Therefore
p(n)
lim
= ad
n→∞ nd
Since ad > 0, Prop 2 concludes that p = Θ(nd ).