   Chapter 11.5, Problem 25ES ### 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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# The recurrence relation for m 1 ,   m 2 ,   m 3 ,   … , which arises in the calculation of the efficiency of merge sort, is   m 1 = 0 m k = m ⌊ k / 2 ⌋ + m ⌊ k / 2 ⌋ + k − 1. Show that for every integer n ≥ 1 ,a. 1 2 n   log 2 n ≥ m n b. m n ≤ 2 n   log 2   n

To determine

(a)

To prove:

For n1, 12nlog2nmn, where the recurrence relation for m1,m2,m3,... is given by

m1=0 and mk=mk2+mk2+k1.

Explanation

Given information:

Recurrence relation for m1,m2,m3,... is,

m1=0mk=m k 2+m k 2+k1.

Concept used:

Mathematical induction:

To prove a statement there are three steps in mathematical induction,

Step 1: Prove that the given statement is true for n=1.

Step 2: Let the given statement is true for n=k, where k is some positive integer greater than 1.

Step 3: In this step, prove that the given statement is true for n=k+1.

Proof:

Given recurrence relation for m1,m2,m3,... is,

m1=0mk=m k 2+m k 2+k1

Consider, the statement as:

12nlog2nmn ……… (1)

Put n=1 in equation (1),

12(1)log2(1)m10=0

So, the statement is true for n=1.

Let the statement is true for n=k, so from equation (1),

12klog2kmk ……….. (2)

Now put n=k+1 in equation (1),

12(k+1)log2(k+1)mk+1 ……….. (3)

Consider left side of equation (2),

12klog2k=12(k+11)log2k=12(k+1)log2k+12(k+1)log2(k+1)12(k+1)log2(k+1)12log2k=12(k+1)log2(k+1)+12(k+1)( log2k log2( k+1))12log2k=12(k+1)log2(k+1

To determine

(b)

To prove:

For n1,mn2nlog2n, where the recurrence relation for m1,m2,m3,..., is given by

m1=0 and mk=mk2+mk2+k1.

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