   Chapter 11.4, Problem 28ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Use Theorems 11.2.7-11.2.9 and properties 11.4.11, 11.4.12, and 11.4.13 to derive each statement in 27-30.28. n 2 + 5 n   log 2 n is Θ ( n 2 )

To determine

Derive the statement: n2+5nlog2n is Θ(n2).

Explanation

Given information n2+5nlog2n

Proof:

n2+5nlog2n

For all real numbers b and r with b > 1 and r > 0, logbnnr for all sufficiently large real numbers x. thus there exists a real number k such that for all real numbers n > k ( b = 2 and r = 1):

log2nn

Note that it is safe to assume that k1. Multiplying both sides of the inequality by 5n, we obtain

5n+log2n5n2

Adding n2 to each side of the inequality, we obtain

n2+5nlog2n6n2

However, n2+5nlog2n is positive when n >1 (as n2 > 0, n > 0 and log2n>0 ) and 6n2 is nonnegative

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