a) Let f: Z+ R where f(n) = ni=1 1. When n = 4, for example,

a) Let f: Z+ †’ R where f(n) = ˆ‘ni=1 1. When n = 4, for example, we have f(n) = f(4) = 1 + 2 + 3 + 4 > 2 + 3 + 4>2 + 2 + 2 = 3- 2 = [(4+ l)/2]2 = 6 > (4/2)2 = (n/2)1. For n = 5, we find f(n) = /(5) = l+2 + 3 + 4 + 5>3 + 4 + 5 > 3 + 3 + 3 = 3ˆ™ 3 = [(5 + 1)(n/2)|3 = 9 > (5/2)2 = (n/2)2. In general, f(n) = 1 + 2 +--------+ n > [n/2] + ˆ™ˆ™ˆ™ˆ™ˆ™ˆ™ˆ™ + n = [n/2] + ˆ™ˆ™ˆ™ˆ™ˆ™ˆ™ + ] = [n/2 = [(n/2)] = [(n + 1)/2][n/2] > n2/4
Consequently, f e Ω(n2).
Use

to provide an alternative proof that f ˆˆ Ω(n2).
(b) Let g: Z+ †’ R where g(n) = ˆ‘ni=1 i2- Prove that g ˆˆ Ω(n3).
(c) For t ˆˆ Z+, let h: Z+ †’ R where h(n) = ˆ‘ni=1 it.
Prove that h ˆˆ Ω(nt+1).

Transcribed Image Text:

-02 n(n 1)