Discrete Mathematics and Applications
Chapter 5: Sequences, Mathematical
Induction and Recursion
5.1 Sequences
Yan Zhang (yzhang16@usf.edu)
Department of Computer Science and Engineering, USF
Sequences
• A sequence is a function.
• Function 𝑓 ∶ 𝐴 → 𝐵
– Every element 𝑥 in 𝐴 is mapped to only one element 𝑦 in 𝐵, such
that (𝑥, 𝑦) ∈ 𝑓.
Domain: A
𝑓∶ 𝐴 → 𝐵
Codomain: B
Not allowed
2
Sequences
Definition
• A sequence is a function.
• Domain is either all the integers between two given integers or all the integers
greater than or equal to a given integer.
Finite sequences: 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2 , ⋯ , 𝑎𝑛
Domain
𝑚
𝑚+1
𝑚+2
⋯
𝑛
Codomain
𝑎𝑚
𝑎𝑚+1
𝑎𝑚+2
⋯
𝑎𝑛
Infinite sequences: 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2 , ⋯
Domain
𝑚
𝑚+1
𝑚+2
⋯
Codomain
𝑎𝑚
𝑎𝑚+1
𝑎𝑚+2
⋯
𝑎𝑘 (read “𝑎 sub 𝑘”) is called a term.
𝑘 in 𝑎𝑘 is called a subscript or index.
Initial term: 𝑎𝑚
Final term: 𝑎𝑛
3
Sequence Example 1
Finding Terms of Sequences Given by Explicit Formulas
Define sequences 𝑎1, 𝑎2, 𝑎3, … and 𝑏2, 𝑏3, 𝑏4, … by the following
explicit formulas:
Compute the first five terms of both sequences.
Solution:
𝑎1 =
𝑏2 =
1
1+1
2−1
2
1
2
1
,
2
2
2
= ,
2+1
3
3−1
2
= ,
3
3
3
3
= ,
3+1
4
4−1
3
= ,
4
4
= , 𝑎2 =
𝑎3 =
𝑎4 =
=
𝑏4 =
𝑏5 =
𝑏3 =
4
4
= ,
4+1
5
5−1
4
= ,
5
5
𝑎5 =
𝑏6 =
5
5
=
5+1
6
6−1
5
=
6
6
Note: all terms of both sequences are identical.
4
Sequence Example 2
Finding an Explicit Formulas to Fit Given Initial Terms
Find an explicit formula for a sequence that has the following
initial terms:
1 1
1 1
1
1, − , , − , , − , ⋯
4 9 16 25 36
Solution:
• Denote the general term of the sequence by 𝑎𝑘 and suppose
the first term is 𝑎1 .
𝑎1 =
1
,
12
𝑎2 = (−1)
1
,
22
𝑎3 =
1
,
32
𝑎4 = (−1)
1
,
42
𝑎5 =
1
,𝑎
52 6
= (−1)
1
62
• An explicit formulator that gives the correct first six terms is:
1
𝑘+1
𝑎𝑘 = (−1)
𝑘2
5
Summation Notation
Given a sequence 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2 , ⋯ , 𝑎𝑛 ,
Domain
𝑚
𝑚+1
𝑚+2
⋯
𝑛
Codomain
𝑎𝑚
𝑎𝑚+1
𝑎𝑚+2
⋯
𝑎𝑛
The summation from 𝑘 equals 𝑚 to 𝑛 of 𝑎-sub-𝑘:
𝑛
𝑘=𝑚 𝑎𝑘
= 𝑎𝑚 + 𝑎𝑚+1 + ⋯ +𝑎𝑛
Summation notation
Expanded form
𝑘: the index of the summation
𝑚: the lower limit of the summation
𝑛: the upper limit of the summation
6
Summation Example
– Computing Summations
Let 𝑎1 = −2, 𝑎2 = −1, 𝑎3 = 0, 𝑎4 = 1, and 𝑎5 = 2.
Compute the following:
a.
5
𝑘=1 𝑎𝑘
5
𝑎𝑘 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5 = −2 + −1 + 0 + 1 + 2 = 0
𝑘=1
b.
c.
2
𝑘=2 𝑎𝑘
2
𝑘=2 𝑎𝑘 = 𝑎2
= −1
2
𝑘=1 𝑎2𝑘
2
𝑘=1 𝑎2𝑘
= 𝑎2 + 𝑎4 = −1 + 1 = 0
7
Summation Example
– Changing from Summation Notation to Expanded Form
Write the following summation in expanded form:
Solution:
8
Summation Example
– Changing from Expanded Form to Summation Notation
Express the following using summation notation:
Solution:
The general term of this summation can be expressed as
for integers k from 0 to n.
Hence
9
Summation Notation - Recursive Definition
If 𝑚 is any integer and 𝑚 < 𝑛, then
𝑚
𝑎𝑘 = 𝑎𝑚
𝑘=𝑚
𝑚+1
𝑚
𝑎𝑘 =
𝑎𝑘 + 𝑎𝑚+1
𝑘=𝑚
𝑘=𝑚
𝑚+2
𝑚+1
𝑎𝑘 =
𝑘=𝑚
𝑛
𝑎𝑘 + 𝑎𝑚+2
𝑘=𝑚
⋯
𝑛−1
𝑎𝑘 =
𝑘=𝑚
𝑎𝑘 + 𝑎𝑛
𝑘=𝑚
Recursive definition is useful to rewrite a summation,
– by separating off the final term of a summation
– by adding a final term to a summation.
10
Recursive Definition Example
– Separating Off a Final Term and Adding On a Final Term
a. Rewrite
b. Write
by separating off the final term.
as a single summation.
Solution:
a.
b.
11
Summation Notation - A Telescoping Sum
• In certain sums each term is a difference of two
quantities.
For example: 𝑎𝑘 =
1
𝑘
1
−
,𝑘
𝑘+1
= 𝑚, ⋯ , 𝑛
• When you write such sums in expanded form, you
sometimes see that all the terms cancel except the
first and the last.
𝑎𝑘 + 𝑎𝑘+1 + 𝑎𝑘+2
1
1
1
1
1
1
=
−
+
−
+
−
𝑘 𝑘+1
𝑘+1 𝑘+2
𝑘+2 𝑘+3
• Successive cancellation of terms collapses the sum
like a telescope.
12
A Telescoping Sum Example
Some sums can be transformed into telescoping sums,
which then can be rewritten as a simple expression.
For instance, observe that
Use this identity to find a simple expression for
13
Product Notation
Given a sequence 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2 , ⋯ , 𝑎𝑛 ,
Domain
𝑚
𝑚+1
𝑚+2
⋯
𝑛
Codomain
𝑎𝑚
𝑎𝑚+1
𝑎𝑚+2
⋯
𝑎𝑛
The product from 𝑘 equals 𝑚 to 𝑛 of 𝑎-sub-𝑘:
𝑛
𝑘=𝑚 𝑎𝑘
= 𝑎𝑚 ∙ 𝑎𝑚+1 ∙ ⋯ ∙ 𝑎𝑛
A recursive definition for the product notation is the following:
If 𝑚 is any integer, then
14
Product Notation Example
– Computing Products
Compute the following products:
a.
b.
Solution:
a.
b.
15
Properties of Summations and Products
16
Properties of Summation & Product Exercise
Let 𝑎𝑘 = 𝑘 + 1 and 𝑏𝑘 = 𝑘 − 1 for all integers 𝑘. Write
each of the following expressions as a single summation or
product:
a.
b.
17
Properties of Summation & Product Exercise
Solution:
a.
18
Properties of Summation & Product Exercise
Solution:
b.
19
Change of Variable
• The index symbol in a summation or product is internal
to summation or product.
• The index symbol can be replaced by any other symbol
as long as the replacement is made in each location
where the symbol occurs.
𝑛
𝑖=𝑚 𝑎𝑖
=
𝑛
𝑗=𝑚 𝑎𝑗
and
𝑛
𝑖=𝑚 𝑎𝑖
=
𝑛
𝑗=𝑚 𝑎𝑗
• As a consequence, the index of a summation or a
product is called a dummy variable.
• A dummy variable is a symbol that derives its entire
meaning from its local context. Outside of that context,
the symbol may have another meaning entirely.
20
Change of Variable Exercise 1
summation:
change of variable:
21
Change of Variable Exercise 1 - Solution
Solution:
• First calculate the lower and upper limits of the new
summation:
Thus the new sum goes from j = 1 to j = 7.
• Next calculate the general term of the new summation by
replacing each occurrence of k by an expression in j :
• Finally, put the steps together to obtain
22
Factorial Notation
A recursive definition for factorial is the following: Given
any nonnegative integer 𝑛,
23
Factorials Exercise
Simplify the following expressions:
a.
b.
c.
d.
Solution:
a.
e.
d.
e.
b.
24
Factorials Exercise Solution
c.
25
“n Choose r ” Notation
Chose 2 objects from {a,b,c,d}:
{a, b}, {a, c}, {a, d}, {b, c}, {b,d}, and {c, d}.
26
“n Choose r” Exercise
Use the formula for computing
expressions:
a.
b.
to evaluate the following
c.
Solution:
a.
27
“n Choose r” Exercise
b.
c.
28
n Choose r Properties
•
•
•
•
•
𝑛
𝑘
𝑛
0
𝑛
𝑘
𝑛
𝑘
𝑛
𝑘
=
=
=
=
=
𝑛
for 0 ≤ 𝑘 ≤ 𝑛
𝑛−𝑘
𝑛
=1
𝑛
𝑛
𝑛−𝑘+1
for 𝑘 > 0
𝑘
𝑘−1
𝑛−1 𝑛
for 𝑘 < 𝑛
𝑛−𝑘
𝑘
𝑛−1 𝑛
for 𝑛, 𝑘 > 0
𝑘−1 𝑘
29
Exercise
Prove that for all nonnegative integers 𝑛 and 𝑟 with 𝑟 + 1 ≤ 𝑛,
𝑛
𝑛−𝑟 𝑛
=
.
𝑟+1 𝑟
𝑟+1
30
Exercise
Prove that for all nonnegative integers 𝑛 and 𝑟 with 𝑟 + 1 ≤ 𝑛,
𝑛
𝑛−𝑟 𝑛
=
.
𝑟+1 𝑟
𝑟+1
Hint:
𝑛−𝑟 𝑛
𝑛−𝑟
𝑛!
=
∙
𝑟
𝑟+1
𝑟 + 1 𝑟! 𝑛 − 𝑟 !
=
𝑛−𝑟
𝑟+1
=
𝑛!
(𝑟+1)! 𝑛−𝑟−1 !
∙
𝑛!
𝑟! 𝑛−𝑟 ∙ 𝑛−𝑟−1 !
𝑛
=
𝑟+1
31