THE UNIVERSITY OF THE WEST INDIES

ASSIGNMENT 2 SEMESTER I, 2012/2013

Code and Name of Course: COMP2101/CS20S – Discrete Mathematics

Assignment 2 - Group Assignment

AREA

DESCRIPTION

Objectives To have students apply the knowledge garnered during weeks 5 to 10

To have students work within a group setting with the allocation of work but

still ensuring that the knowledge is spread throughout the group

Title Assignment 2 – Group Assignment

Deliverable The answers for the questions which follow the given instructions.

A summary of the contribution made by each student in the group.

Instructions 1. Review Lectures of weeks 5 to 10 of the course, the COMP2101 Text and

any other related Discrete Mathematics material

2. Read the Assignment 2 Sheet thoroughly

3. Submit the gradable solution by using the

COMP2101 Assignment 2 - Group located within the ASSIGNMENTS

Section of the OURVLE COMP2101/CS20S Course Environment. This

may also be accessed by choosing the Assignments section below

“Activities”

Format The solution for this assignment must be submitted as a Microsoft Word

document.

Your ID number should form part of the Microsoft Word file name. The file

name should take the format “COMP2101 Assignment 2 Semester 1 2012-

2013 XXXXXXXX YYYYYYYY ZZZZZZZZ” where XXXXXXXX, etc.

represents the student ID numbers.

Upload

Constraint

Assignment should be uploaded in the relevant space provided in

OURVLE (See “Instructions” section above). A message indicating “File

uploaded successfully” will acknowledge that the file has been sent

successfully.

Do NOT assume your project has been received if you do not get this

acknowledgement.

Group Restrictions

This is a group assignment. Each group must have a minimum of 2 members and a maximum of four (4) members.

Late

Assignments

Late assignments are accepted. These are however graded then 25%

deducted for each day of late submission.

Expectation It is expected that students will discuss means to a solution within their group.

The actual work written is expected to reflect the group’s decision. Where

replication of work is identified between groups, each paper will be graded.

The allocated grade to each group’s piece of work for where this anomaly is

identified will be the grade divided by the number of replications discovered.

Scoring Rubric

Your electronic submission will be evaluated on:

1. The response submitted for each question (See detail individual marks

below) – 45 marks

2. Group Dynamics i.e. for accomplishing the task as a group – 5 marks.

Marks allocated based on correct inclusion of each student id number and

name and a brief description as to the contribution made within the group.

The actual grade of 50 marks will be displayed. The actual grade allocated is

the percentage of the maximum marks (5 points)

Due Date

Monday, November 19, 2012

Question 1 [8 mks]

A coding system encodes messages using strings of octal digits (base 8). A codeword is

considered valid if and only if it contains an even number of 7s.

i. Find a recurrence relation for the number of valid codewords of length n.

State initial conditions. [4]

ii. Solve this recurrence relation using generating functions. [4]

Question 2 [2 mks]

a. Determine the limit of f(x) as x for the following:

33

5)(

x

xxf [1]

b. For the sequence a and b defined by

a1 = 2, an = 3an – 1, n ≥ 2, and bn = 2n (n – 1)

Find ))((7

3

5

2

j

j

i

i ba [1]

Question 3 [5 mks]

Find a formula for the sequence c defined by

n

i

in bc1

Where b is the sequence 1, 2, 4, 4, 7, 8, 10, … [5]

Question 4 [3 mks]

Let f(n) and g(n) be functions defined on the set of positive integers

Prove or disprove the following:

if f (n) = Ο (h(n)) and g(n) = Ο ( k(n))

then f(n)g(n) = Ο (h(n)k(n)). [3]

Question 5 [2 mks]

What is the limit of

n

k

k nasx0

? State all necessary

conditions for this limit to exist. [2]

Question 6 [7 mks]

Use generating functions to find a closed form solution for each of the following

recurrence relations:

i. a0 = 1 and an = 3an-1 + 2 for n ≥ 1 [3]

ii. s0 = 6, s1 = 5 and 9Sn = 6Sn-1 - Sn-2 for n ≥ 0 [4]

Question 7 [8 mks]

Let a,b,c be integers such that a ≥ 1, b > 1 and c > 0. Let f: N → R be functions

where N is the set of Natural numbers and R is the set of Real numbers

such that

f(1) = c and f(n) = af(n/b) + c

for n = bk, where k is a positive integer greater than 1.

(a) By using the principles of Recurrence Relation, find a general formula

for f(n) [4]

(b) Hence show that if a ≠ 1, then f(n) = )(1

)1( loglog

aa

b

b

na

anc

[4]

Question 8 [4 mks]

Consider the geometric series:

3 + 12/7 + 48/49 + 192/343 + …

i. Determine a formula for Sn

where Sn is the sum of the first n terms of the series? [3]

ii. What is the limit of Sn [1]

Question 9 [6 mks]

Find the generating function for the sequence

1, 4, 13, 28, 49, … [6]

Question 1 [8 mks]

A coding system encodes messages using strings of octal digits (base 8). A codeword is

considered valid if and only if it contains an even number of 7s.

i. Find a recurrence relation for the number of valid codewords of length n.

State initial conditions. [4]

ii. Solve this recurrence relation using generating functions. [4]

Solution 1

[i. Logical steps outlined towards finding general equation - 3 marks ]

[ Finding general equation f(n) or Sn - 1 mark ]

[ii. Correct initial layout of solution – 1st 3 statements - 1 mark ]

[ Application of Correct combining i.e. Subtraction - 1 mark ]

[ Logical steps towards correct solution - 2 marks ]

i.

Solution A - IF ZERO 7’S IS NOT CONSIDERED TO BE EVEN

Set of strings of 0, 1, 2, 3, 4, 5, 6, 7

Valid – String contains an even number of 7s e.g. 765703377112 or 001767 or 77

Let n be the length of the codeword

Sn be the number of valid codewords of length n or

Sn is the number of codewords of length n with an even number of 7s

i.e. Sn-1 is the number of codewords of length n-1 with an even number of 7s

Sn-2 is the number of codewords of length n-2 with an even number of 7s

etc.

By Counting

S0 = 0

S1 = 0

S2 = 1

S3 = 21

.

.

Firstly,

The valid codewords begins with a ‘7’ or Not

Let SA be the number of valid codewords of length n that Do Not begin with 7

SB be the number of valid codewords of length n that begins with 7

Then

Sn = SA + SB

Consider SA [Does Not Begin with ‘7’, i.e. begins with 0 or 1 or ... or 6]

Let us say a valid codeword that begins with 0

The other n-1 octal digits must contain an even number of 7s i.e. Sn-1

As there are 7 such cases

SA = 7Sn-1

Consider SB [Begins with ‘7’]

The other n-1 octal digits must contain an odd number of 7s

For these n-1 octal digits

The number of odds = Total number of codewords

minus The number of evens

minus The number without 7s (not odd or even)

Therefore,

As we are now considering the n-1 octal digits

SB = 8n-1

- Sn-1 - 7n-1

Therefore, As

Sn = SA + SB

The Recurrence Relation

Sn = 7Sn-1 + 8n-1

- Sn-1 - 7n-1

Simplified

Sn = 6Sn-1 + 8n-1

- 7n-1

for n > 1

For The Initial Conditions

When n = 1,

S1 = 0

S1 = 6S0 + 80 - 7

0

Therefore

S0 = 0

Solution B - IF ZERO 7’S IS CONSIDERED TO BE EVEN

Set of strings of 0, 1, 2, 3, 4, 5, 6, 7

Valid – String contains an even number of 7s e.g. 765703377112 or 001767 or 77

or 0123456065 or 333333 or 00 or 5

Let n be the length of the codeword

Sn be the number of valid codewords of length n or

Sn is the number of codewords of length n with an even number of 7s

i.e. Sn-1 is the number of codewords of length n-1 with an even number of 7s

Sn-2 is the number of codewords of length n-2 with an even number of 7s

etc.

By Counting

S1 = 7

S2 = 72 + 1 = 50

S3 = 73 + 3x7 = 364

.

.

Firstly,

The valid codewords begins with a ‘7’ or Not

Let SA be the number of valid codewords of length n that Do Not begin with 7

SB be the number of valid codewords of length n that begins with 7

Then

Sn = SA + SB

Consider SA [Does Not Begin with ‘7’, i.e. begins with 0 or 1 or ... or 6]

Let us say a valid codeword that begins with 0

The other n-1 octal digits must contain an even number of 7s i.e. Sn-1

As there are 7 such cases

SA = 7Sn-1

Consider SB [Begins with ‘7’]

The other n-1 octal digits must contain an odd number of 7s

For these n-1 octal digits

The number of odds = Total number of codewords

minus The number of evens

minus The number without 7s (not odd or even)

Therefore,

As we are now considering the n-1 octal digits

SB = 8n-1

- Sn-1

Therefore, As

Sn = SA + SB

The Recurrence Relation

Sn = 7Sn-1 + 8n-1

- Sn-1

Simplified

Sn = 6Sn-1 + 8n-1

for n > 1

For The Initial Conditions

When n = 1,

S1 = 7

S1 = 6S0 + 80

Therefore

S0 = 1

ii.

Solution A - IF ZERO 7’S IS NOT CONSIDERED TO BE EVEN

S0 = 0

S1 = 0

Sn = 6Sn-1 + 8n-1

- 7n-1

for n > 1

Let S = s0 + s1x + s2x2 + s3x

3 + ... + snx

n + ...

6xS = 6s0x + 6s1x2 + 6s2x

3+ ... + 6sn-1x

n + ...

It follows

Sn - 6Sn-1 = 8n-1

- 7n-1

for n > 1

By Subtraction

S(1- 6x) = s0 + (s1 - 6s0)x + (s2-6s1)x2 + (s3-6s2)x

3 +... + (sn-6sn-1)x

n + ...

S2 = 1

S3 = 6 + 64 – 49 = 21

S(1- 6x) = (81 -7

1)x

2 +(8

2 -7

2)x

3 +(8

3 -7

3)x

4 +...+ (8

n-1 -7

n-1)x

n +...

S = [ (81 -7

1)x

2 +(8

2 -7

2)x

3 +(8

3 -7

3)x

4 +...+ (8

n-1 -7

n-1)x

n +... ] / (1-6x)

As 1/(1-6x) = 1 + 6x + 62x

2 + 6

3x

3 + ... + 6

nx

n + ...

S = [(81-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] *

(1+6x+62x

2+6

3x

3+ .. + 6

nx

n+ ...)

S = 1 [(81-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] +

6x [(81-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] +

62x

2 [(8

1-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] +

63x

3 [(8

1-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] +

.

.

6nx

n [(8

1-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] +

...

= (8-7)x2 + (8

2-7

2+6(8-7))x

3 + (8

3-7

3+6(8

2-7

2)+6

2(8-7))x

4 +

(84-7

4+6(8

3-7

3) +6

2(8

2-7

2) +6

3(8-7))x

5 + ... +

[8n-1

-7n-1

+6(8n-2

-7n-2

) +62(8

n-3-7

n-3)+...+6

n-3(8

2-7

2)+6

n-2(8-7) ]x

n +...

Therefore (before simplification) the closed form solution of the recurrence relation is

[ 8n-1

-7n-1

+6(8n-2

-7n-2

) +62(8

n-3-7

n-3)+...+6

n-3(8

2-7

2)+6

n-2(8-7) ]

Simplified

[ 60(8

n-1-7

n-1)+6(8

n-2-7

n-2) +6

2(8

n-3-7

n-3)+...+6

n-3(8

2-7

2)+6

n-2(8-7) ]

2

0

1212 )78(6n

k

knknk

2

0

11 )78(6n

k

knknk

The closed form solution of the recurrence relation is

2

0

11 )78(6n

k

knknk

Solution B - IF ZERO 7’S IS CONSIDERED TO BE EVEN

S0 = 1

S1 = 7

Sn = 6Sn-1 + 8n-1

for n > 1

Let S = s0 + s1x + s2x2 + s3x

3 + ... + snx

n + ...

6xS = 6s0x + 6s1x2 + 6s2x

3+ ... + 6sn-1x

n + ...

It follows

Sn - 6Sn-1 = 8n-1

for n > 1

By Subtraction

S(1- 6x) = s0 + (s1 - 6s0)x + (s2-6s1)x2 + (s3-6s2)x

3 +... + (sn-6sn-1)x

n + ...

S2 = 72 + 1 = 50

S3 = 73 + 3x7 = 364

S(1- 6x) = 1 + x + 81x

2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n +...

S = [1 + x + 8x2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ... ] / (1-6x)

As 1/(1-6x) = 1 + 6x + 62x

2 + 6

3x

3 + ... + 6

nx

n + ...

S = [(81-7

1)x

2+(8

2-7

2)x

3+(8

3-7

3)x

4+..+(8

n-1-7

n-1)x

n+..] *

(1+6x+62x

2+6

3x

3+ .. + 6

nx

n+ ...)

S = 1 [1 + x + 8x2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ...] +

6x [1 + x + 8x2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ...] +

62x

2 [1 + x + 8x

2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ...+

63x

3 [1 + x + 8x

2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ...+

.

.

6nx

n [(1 + x + 8x

2 + 8

2x

3 + 8

3x

4 + ... + 8

n-1x

n + ...] +

...

= 1 + (1+6)x + (8+6+62)x

2 + (8

2+6(8) +6

2+6

3)x

3 +

(83+6(8

2)+6

2(8)+ 6

3 +6

4)x

4 +

(84+6(8

3) +6

2(8

2) +6

3(8) + 6

4 +6

5)x

5 +

... +

[8n-1

+6(8n-2

) +62(8

n-3)+...+6

n-3(8

2)+6

n-2(8) + 6

n-1 +6

n ]x

n +...

Therefore (before simplification) the closed form solution of the recurrence relation is

[ 8n-1

+6(8n-2

) +62(8

n-3)+...+6

n-3(8

2)+6

n-2(8) + 6

n-1 +6

n ]

Simplified

[ 608

n-1+6

1 (8

n-2) +6

2(8

n-3)+...+6

n-3(8

2)+6

n-2(8

1) + 6

n-1(8

0) +6

n ]

nn

k

knk 6)8(61

0

1

The closed form solution of the recurrence relation is

1

0

1 )8(66n

k

knkn

Question 2 [2 mks]

a. Determine the limit of f(x) as x for the following:

33

5)(

x

xxf [1]

b. For the sequence a and b defined by

a1 = 2, an = 3an – 1, n ≥ 2, and bn = 2n (n – 1)

Find ))((7

3

5

2

j

j

i

i ba [1]

Solution 2

[a. Correct solution - 1 mark ]

[ Where question is attempted but incorrect - ½ mark ]

[b. Correct solution - 1 mark ]

[ Where question is attempted but incorrect - ½ mark ]

a. 33

5)(

x

xxf

Dividing numerator and denominator by x, we have

x

xf3

3

5)(

As x →∞, x

3→0

3

5

33

5)( limlim

x

xxf

xx

b. ))((7

3

5

2

j

j

i

i ba = (a2 + a3 + a4 + a5) * (b3 * b4 * b5 * b6 * b7 )

= (6 + 18 + 54 + 162) * (16 * 48 * 128 * 320 * 768)

= 240 * 24159191040

= 5,798,205,849,600

Question 3 [5 mks]

Find a formula for the sequence c defined by

n

i

in bc1

Where b is the sequence 1, 2, 4, 4, 7, 8, 10, … [5]

Solution 3

[For stating an Alternating AP-GP Sequence - 1 mark ]

[First terms a, Common difference d & ratio r, formulae for Sn - 1 mark ]

[Solving Sn to produce correct formulae - 2 marks ]

[Correct combination of AP and GP formulae to produce cn - 1 mark ]

For the sequence b defined by 1, 2, 4, 4, 7, 8, 10, …

b1 = 1,

b2 = 2,

b3 = 4,

b4 = 4

b5 = 7,

b6 = 8,

b7 = 10,

b8 = 16 (assumed)

For odd number indexes in the sequence, we have an Arithmetic Progression, with

a = 1

d = 3

For even number indexes in the sequence, we have a Geometric Progression, with

a = 2

r = 2

We have Alternating AP-GP sequence

For the AP sequence

Sn = (n/2)(2a + (n-1)d)

= (n/2)(2x1 + (n-1)x3)

= (n/2)(2 + 3n - 3)

= (n/2)(3n - 1)

As in our AP formula c1 represents S1 for the odd number indexes

c1 represents S1

c3 represents S2

c5 represents S3 and so on

i.e. cn represents S(n+1)/2

For odd number indexes

Sn = ( n /2)(3n - 1)

By substitution

cn = ( ((n+1)/2) /2)(3((n+1)/2) - 1)

= ( (n+1)/4)((3n+3)/2 - 1)

= ( n/4 + 1/4 )( 3n/2 + ½ )

= ½ ¼ ( n + 1 )( 3n + 1 )

= (1/8)( n + 1 )( 3n + 1 )

For the GP sequence

Sn = a(rn - 1)/(r – 1)

= 2(2n - 1)/(2 – 1)

= 2(2n - 1)

As in our GP formula c1 represents S1 for the even number indexes

c2 represents S1

c4 represents S2

c6 represents S3 and so on

i.e. cn represents Sn/2

For even number indexes

Sn = 2 (2n - 1)

By substitution

cn = 2 (2n/2

- 1)

When n is odd,

cn = cn (for odd) + cn-1 (for even)

= (1/8)( n + 1 )( 3n + 1 ) + 2(2(n-1)/2

- 1)

When n is even,

cn = cn (for even) + cn-1 (for odd)

= 2(2n/2

- 1) + (1/8)( (n-1) + 1 )( 3(n-1) + 1 )

= 2(2n/2

- 1) + (1/8)( n )( 3n - 3 + 1 )

= 2(2n/2

- 1) + (1/8)( n )( 3n - 2 )

Question 4 [3 mks]

Let f(n) and g(n) be functions defined on the set of positive integers

Prove or disprove the following:

if f (n) = Ο (h(n)) and g(n) = Ο ( k(n))

then f(n)g(n) = Ο (h(n)k(n)). [3]

Solution 4

[Stating f(n) and g(n) in terms of the inequality - 1 mark ]

[Stating that | f(n) | | g(n) | = | f(n) g(n) | for positive integers - ½ mark ]

[Logical steps of the Proof - 1½ marks ]

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

⇒ | f(n) | ≤ C1 | h(n) |

where C1 is a constant

g(n) = O(k(n))

⇒ | g(n) | ≤ C2 | k(n) |

where C2 is a constant

By Multiplication

| f(n) | | g(n) | ≤ C1 C2 | h(n) | | k(n) |

As f(n) and g(n) are functions defined on the set of positive integers

| f(n) | | g(n) | = | f(n) g(n) | ≤ C1 C2 | h(n) | | k(n) |

∴ | f(n) g(n) | ≤ C1 C2 | h(n) | | k(n) |

| f(n) g(n) | ≤ C3 | h(n) | | k(n) |

where C3 is a new constant

Where h(n) and k(n) are functions defined on the set of positive integers

| f(n) g(n) | ≤ C3 | h(n) | | k(n) | = C3 | h(n) k(n) |

∴ | f(n) g(n) | ≤ C3 | h(n) k(n) |

∴ f(n) g(n) = O(h(n) k(n))

Question 5 [2 mks]

What is the limit of

n

k

k nasx0

? State all necessary

conditions for this limit to exist. [2]

Solution 5

[Correct limit found - 1½ marks ]

[Necessary conditions - ½ mark ]

As n →∞

n

k

kx0

= 1 + x + x 2 + x

3 +…

By Generating Functions definition, nas

1 + x + x 2

+ x 3 +… = 1 / (1 - x)

Condition: x ≠ 1

Question 6 [7 mks]

Use generating functions to find a closed form solution for each of the following

recurrence relations:

i. a0 = 1 and an = 3an-1 + 2 for n ≥ 1 [3]

ii. s0 = 6, s1 = 5 and 9Sn = 6Sn-1 - Sn-2 for n ≥ 0 [4]

Solution 6

[i. Correct initial layout of solution – 1st 2 statements - ½ mark ]

[ Application of Correct combining i.e. Subtraction - ½ mark ]

[ Logical steps towards correct solution - 2 marks]

[ Where full mark is not attained, but there was simplification of solution included ]

[ additional ½ mark ]

[ii. Correct initial layout of solution – 1st 3 statements - 1 mark ]

[ Application of Correct combining i.e. Subtraction & Addition - 1 mark ]

[ Logical steps towards correct solution - 3 marks ]

[ Where full mark is not attained, but there was simplification of solution included ]

[ additional ½ mark ]

i. Given

a0 = 1

an = 3an-1 + 2 , n≥1

Consider the generating function

f(x) = a0+a1x+ a2x2+…+ anx

n+…

3xf(x) = 3a0x+3a1x2+…+3an-1x

n+…

It follows

an - 3an-1 = 2

Subtracting…

f(x) - 3xf(x) = a0 + (a1- 3a0)x + (a2- 3a1)x2+…+ (an- 3an-1)x

n +…

Now Substituting a0 = 1, a1 = 3a0 ,…, an = 3an-1

(1-3x)f(x) = 1 + 2x + 2x2+…+ 2x

n +…

= -1 + 2 (1 + x + x2+…+ x

n +…)

f(x) = (-1 + 2 (1 + x + x2+…+ x

n +…)) * (1 / (1-3x))

f(x) = (-1 + 2 (1 + x + x2+…+ x

n +…)) * (1 + 3x + 3

2x

2+…+ 3

nx

n +…)

f(x) = -1 * (1 + 3x + 32x

2+…+ 3

nx

n +…)

+ 2 [ 1 * (1 + 3x + 32x

2+…+ 3

nx

n +…)

+ x * (1 + 3x + 32x

2+…+ 3

nx

n +…)

+ x2 * (1 + 3x + 3

2x

2+…+ 3

nx

n +…

…

+ xn * (1 + 3x + 3

2x

2+…+ 3

nx

n +…) ]

f(x) = -1 * (1 + 3x + 32x

2+…+ 3

nx

n +…)

+ 2 [ 1 + (3 + 1)x + (32+3+1)x

2 + …+ (3

n+3

n-1+…+3

2+3+1) x

n +… ]

Therefore the closed form solution is

[ -3n + 2 (3

n+3

n-1+…+3

2+3+1) ]

or

n

k

kn

0

323

ii. S0 = 6

S1 = 5 and

9Sn = 6Sn-1 - Sn-2 for n ≥ 2

Let S = s0 + s1x + s2x2 + s3x

3 + ... + snx

n + ...

9S = 9s0 + 9s1x + 9s2x2 + 9s3x

3 + ... + 9snx

n + ...

6xS = 6s0x + 6s1x2 + 6s2x

3+ . . . + 6sn-1x

n + ...

x2S = s0x

2 + s1x

3+ . . . + sn-2x

n + ...

9Sn - 6Sn-1 + Sn-2 = 0

By Subtraction and Addition

S(9- 6x+x2) = 9s0 + (9s1 - 6s0)x + (9s2-6s1+s0)x

2 + ... + (9sn-6sn-1 + sn-2)x

n + ...

= 54 + 9x

Dividing through by 9

S(1- (2/3)x+(1/9)x2) = 6 + x

S(1-(1/3)x) (1-(1/3)x) = 6 + x

S = (6 + x) / [(1-(1/3)x)(1-(1/3)x)] or (6 + x) / [(1-(1/3)x)]2

As 1/(1-(1/3)x) = 1 + (1/3)x + (1/3)2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...

S = (6+x)(1+(1/3)x+(1/3)2x

2+ ...+(1/3)

nx

n + ...) (1+(1/3)x+(1/3)

2x

2+ ... +(1/3)

nx

n + ...)

= (6+x) * [

1 (1 + (1/3)x + (1/3)2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...) +

(1/3)x (1 + (1/3)x + (1/3)2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...) +

(1/3)2x

2 (1 + (1/3)x + (1/3)

2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...) +

(1/3)3x

3 (1 + (1/3)x + (1/3)

2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...) +

.

.

(1/3)nx

n (1 + (1/3)x + (1/3)

2x

2 + (1/3)

3x

3 + ... + (1/3)

nx

n + ...)+ ]

= (6+x) [1+2(1/3)x + 3(1/3)2x

2 + 4(1/3)

3x

3+...+ n(1/3)

n-1x

n-1+(n+1)(1/3)

nx

n+...]

= 6 * [1+2(1/3)x + 3(1/3)2x

2 + 4(1/3)

3x

3+...+ n(1/3)

n-1x

n-1+(n+1)(1/3)

nx

n+...]

+ x [1+2(1/3)x + 3(1/3)2x

2 + 4(1/3)

3x

3+...+ n(1/3)

n-1x

n-1+(n+1)(1/3)

nx

n+...]

= 6 + [6*2(1/3)+1]x + [6*3(1/3)2+2(1/3)]x

2 + [6*4(1/3)

3+3(1/3)

2]x

3+

+ [6*5(1/3)4+4(1/3)

3]x

4+...

+ [6*(n+1)(1/3)n+n(1/3)

n-1]x

n+...

Therefore (before simplification) the closed form solution of the recurrence relation is

[6 * (n+1) (1/3)n

+ n (1/3)n-1

]

Simplified

[6 * (n+1) (1/3)n-1+1

+ n (1/3)n-1

]

[(1/3)n-1

(6 * (n+1) (1/3)1

+ n) ]

[(1/3)n-1

(6 * (1/3)1 (n+1)

+ n) ]

[(1/3)n-1

(2 * (n+1) + n) ]

[(1/3)n-1

(3n+2) ]

The closed form solution of the recurrence relation is

[ (1/3)n-1

(3n+2) ]

or [ (1/3)n(1/3)

-1 (3n+2) ]

or [ (1/3)n(3)

(3n+2) ]

or [ (1/3)n(9n+6) ]

Question 7 [8 mks]

Let a,b,c be integers such that a ≥ 1, b > 1 and c > 0. Let f: N → R be functions

where N is the set of Natural numbers and R is the set of Real numbers

such that

f(1) = c and f(n) = af(n/b) + c

for n = bk, where k is a positive integer greater than 1.

(a) By using the principles of Recurrence Relation, find a general formula

for f(n) [4]

(b) Hence show that if a ≠ 1, then f(n) = )(1

)1( loglog

aa

b

b

na

anc

[4]

Solution 7

[(a) Logical steps outlined towards finding general equation - 3 marks ]

[ Finding general equation f(n) - 1 mark ]

[(b) Making substitutions and finding f(n) = calog

bn + c

1log

0

n

i

ib

a - 1 mark ]

[ Proving Big-Oh - 2 marks ]

[ Proving Omega - 1 mark ]

[Where error is made in any of the above but proofs for Big-Oh and Omega were

attempted and full mark is not attained - ½ mark ]

(a) f(n) = a f(n/b) + c .....1

Substituting for f(n/b)

= a [ a f(n/b2) + c ] + c

= a2 f(n/b

2) + c(a + 1) .....2

Substituting for f(n/b2)

= a2 [ a f(n/b

3 ) + c ] + c(a + 1)

= a3 f(n/b

3 ) + c(a

2 + a + 1) .....3

Substituting for f(n/b3)

= a3 [ a f(n/ b

4) + c ] + c(a

2 + a + 1)

= a4 f(n/b

4 ) + c(a

3 + a

2 + a + 1) .....4

= ......

Given that k is a positive integer greater than 1

f(n) = ak f(n/b

k) + c

1

0

k

i

ia

(b) As n = bk

n/bk = 1 AND logb n = k

As k = logb n AND n/bk = 1 AND if a ≠ 1 AND f(1) = c

f(n) = ak f(n/b

k) + c

1

0

k

i

ia

= ak f(1) + c

1

0

k

i

ia

= ak * f(1) + c * [a

k-1 + a

k-2 +…+ a

3 + a

2 + a + 1]

As f(1) = c, and k = logb n

f(n) = calog

bn + c

1log

0

n

i

ib

a

f(n) = calog

bn + c [a

logb

n-1 + a

logb

n-2 +…+ a

3 + a

2 + a + 1]

= c[alog

bn + a

logb

n-1 + a

logb

n-2 +…+ a

3 + a

2 + a + 1]

Further, as logb n = k AND k is a positive integer greater than 1

f(n) ≤ c[alog

bn + a

logb

n + a

logb

n +…+ a

logb

n + a

logb

n + a

logb

n + a

logb

n]

≤ c(n+1)alog

bn

Therefore

f(n) = O(alog

b n

)

Likewise

f(n) = calog

bn + c

1log

0

n

i

ib

a

f(n) ≥ calog

bn

Therefore

f(n) = Ω(alog

b n

)

Hence

f(n) = Θ(alog

b n

)

Question 8 [4 mks]

Consider the geometric series:

3 + 12/7 + 48/49 + 192/343 + …

i. Determine a formula for Sn

where Sn is the sum of the first n terms of the series? [3]

ii. What is the limit of Sn [1]

Solution 8

[i. Common ratio, r - ½ mark ]

[ Correct formula for Sn - ½ mark ]

[ Solving Sn to produce a correct formula - 2 marks ]

[ii. Correct formula for |r| < 1 - ½ mark ]

[ Correct solution for limit of Sn - ½ mark ]

i. For a geometric progression

un = arn-1

u1 = a = 3

u2 = ar = 12/7

r = ar / a = (12/7) / 3

= 12/21

= 4/7

Sn = a(rn - 1)/(r - 1)

Sn = 3((4/7)n - 1) / (4/7 - 1)

= 3((4/7)n - 1) / -3/7

= 3(1 - (4/7)n) / 3/7

= 3(1 - (4/7)n) * 7/3

= 7(1 - (4/7)n)

or

7 - 7(4/7)n

or

-7( (4/7)n - 1)

or

-7(4/7)n + 7

ii. Where |r| < 1

Limit of Sn = a / (1 – r)

= 3 / (1 – 4/7)

= 3 / (3/7)

= 7

Question 9 [6 mks]

Find the generating function for the sequence

1, 4, 13, 28, 49, … [6]

Solution 9

[Logical steps towards correct solution - 3 marks ]

[Correct interpretation of operations and computation - 2 marks ]

[Correct Final Solution - 1 mark ]

We have 1, 1, 1, 1, … ≡ 1 + x + x2 + x

3 + x

4 + …≡

x1

1

By Differentiating

1, 2, 3, 4,5 … ≡ 2)1(

1

x

By Right-shifting, 1 place

0, 1, 2, 3, 4, … ≡ 2)1( x

x

By Differentiating, again

For LHS

1, 4, 9, 16, …

For RHS

2

2)1(

)1(

xxx

xyLet

dx

dp

dp

dvU

dx

duV

dx

dythen

dx

dp

dp

dv

dx

dvand

pvxpLet

Further

dx

dvU

dx

duV

dx

dythen

vuyxvxuLet

2

2

,1

,)1(,

3

3

32

32

32

22

)1(

1

)1(

21

)1(

2

)1(

1

)1(2)1(

)1()2()1()1(

)1()()1(

x

x

x

xx

x

x

x

xxx

pxx

dx

xd

dp

pdx

dx

dxx

dx

dy

Therefore

1, 4, 9, 16, 25,… ≡ 3)1(

1

x

x

By Right-shifting, 1 place

0, 1, 4, 9, 16,,… ≡ 3)1(

)1(

x

xx

By Scaling, with factor of 3

0, 3, 12, 27, 48, … ≡ 3)1(

)1(3

x

xx

By Addition of generating function 1, 1, 1, 1, ...

For LHS

1, 4, 13, 28, 49, …

For RHS

3

2

3

22

3

2

3

)1(

14

)1(

2133

)1(

)1()1(3

1

1

)1(

)1(3

x

xx

x

xxxx

x

xxx

xx

xx

1, 4, 13, 28, 49, … ≡ 3

2

)1(

14

x

xx

Therefore the generating function for the sequence 1, 4, 13, 28, 49, … is 3

2

)1(

14

x

xx