5장 Sequences and Mathematical Induction
Sequences
Definition
A sequence is a function whose domain is either all the integers between two given integers or
all the integers greater than or equal to a given integer.
A set of (usually infinite number of) ordered elements written in a row.
a0, a1, a2, a3, …, an
- Each individual elements an ( read “a sub n” ) is called a Term
- n in an is called a subscript or index.
a0 (initial Term), an (final Term)
- Explicit formula or General formula for a sequence
a Rule that shows how the values of an depend on n.
Sequences
Example 5.1.1 Finding Terms of Sequences Given by Explicit Formulas
Define sequences a1, a2, a3, … and b1, b2, b3, … by the following explict formulas:
k
,
k 1
i 1
bi 
,
i
ak 
k 1
i  2
Compute the first five terms of both sequences.
Sequences
Example 5.1.1 Finding Terms of Sequences Given by Explicit Formulas
Define sequences a1, a2, a3, … and b1, b2, b3, … by the following explict formulas:
k
,
k 1
i 1
bi 
,
i
ak 
k 1
i  2
Compute the first five terms of both sequences.
Sequences
Example 5.1.2 An Alternating Sequence
Compute the first six terms of sequences a1, a2, a3, … defined as follows:
an  (1) n for every integer n ≥ 0
a0  ( 1) 0 
1
a3  ( 1) 3  1
a1  ( 1)1  1
a4  ( 1) 4 
a2  ( 1) 2 
a5  ( 1) 5  1
1
1
Sequences
Example 5.1.3 Finding an Explicit Formula to Fit Given Initial Terms
Find an explicit formula for a sequence with the following initial terms:
Sequences
Example 5.1.3 Finding an Explicit Formula to Fit Given Initial Terms
Find an explicit formula for a sequence with the following initial terms:
1, -1, 1, -1,…
When k is odd, k + 1 is even and thus
when k is even, k + 1 is odd and thus
Thus, an explicit formula is
If the first term is a0, then the explicit formula is
Sequences
Definition of Summation
𝑛
If m and n are integers and m ≤ n, the symbol
𝑎𝑘
, read the summation from k
𝑘=𝑚
equals m to n of a-sub-k, is the sum of all the terms 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2, …, 𝑎𝑛 .
We say that 𝑎𝑚 + 𝑎𝑚+1 + 𝑎𝑚+2 + …+ 𝑎𝑛 is the expanded form of the sum, and we write
𝑛
𝑎𝑘
= 𝑎𝑚 + 𝑎𝑚+1 + 𝑎𝑚+2 + … + 𝑎𝑛
𝑘=𝑚
We call k the index of the summation, m the lower limit of the summation, and
n the upper limit of the summation.
정리:
Summation from k equals m to n
－ k : index of summation
－ m : lower limit of summation
－ n : upper limit
Sequences
Example 5.1.4 Computing Summations
Let 𝑎1 = -2, 𝑎2 = -1, 𝑎3 = 0, 𝑎4 = 1, and 𝑎5 = 2. Computing following:
a.
5
𝑎𝑘
= 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5 = (-2) + (-1) + 0 + 1 + 2 = 0
𝑎𝑘
= 𝑎2 = -1
𝑘=1
b.
2
𝑘=2
c.
2
𝑎2𝑘
𝑘=1
= 𝑎2∙1 + 𝑎2∙2 = 𝑎2 + 𝑎4 = -1 + 1 = 0
Example 5.1.5 When the Terms of a Summation Are Given by a Formula
Compute
Sequences
Example 5.1.6 Changing from Summation Notation to Expanded Form
Write
in expanded form:
Example 5.1.7 Changing from Expanded Form to Summation Notation
Express the following using summation notation:
Sequences
Recursive Definition
If m and n are any integers with m < n, then
m
a
k m
k
n 1
n
a
 am and
k m
k
 an   ak for all integers n > m
k m
either by separating off the final term of a summation or adding a final term to a summation.
Example 5.1.9 Using a Single Summation Sign and Separating Off a Final Term
a. Write
n 1
2
k
 2 n as a single summation.
k 0
b. Rewrite
n
1
i
i 0
2
by separating off the final term.
Sequences
Example 5.1.10 A Telescoping Sum
Some sums can be transformed into Telescoping sums
n
1

k 1 k ( k  1)
Sequences
Definition of Product Notation
If m and n are integers and m ≤ n, the symbol
𝑛
𝑎𝑘,
read the product from k
𝑘=𝑚
equals m to n of a-sub-k, is the product of all the terms 𝑎𝑚 , 𝑎𝑚+1, 𝑎𝑚+2 , …, 𝑎𝑛 .
We write
𝑛
𝑎𝑘
𝑘=𝑚
= 𝑎𝑚 ∙𝑎𝑚+1 ∙𝑎𝑚 ∙, …, ∙𝑎𝑚
Recursive Definition
If m and n are any integers with m < n, then
𝑚
𝑎𝑘
= 𝑎𝑚 and
𝑘=𝑚
𝑛
𝑎𝑘
𝑘=𝑚
𝑛
=
𝑎𝑘
∙𝑎𝑛
𝑘=𝑚
Example 5.1.11 Computing Products
Compute the following products:
a.
5
 k  1  2  3  4  5  120
k 1
1
b.
k
1
1
 k 1  11  2
k 1
for all integers n > m
Properties of Summations and Products
Theorem 5.1.1
if 𝑎𝑚 , 𝑎𝑚+1, 𝑎𝑚+2 , …, and 𝑏𝑚 , 𝑏𝑚+1, 𝑏𝑚+2 , … are sequence of real numbers and
c is any real number, then the following equations hold for any integer n ≥ m.
1.
2.
3.
n
n
n
 a   b   (a
k m
k
k m
k
k m
n
n
k m
k m
k
 bk )
c   ak   c  ak
 n
  n

  ak     bk  
 k m   k m 
generalized distributive law
n
 (a
k m
k
 bk )
Properties of Summations and Products
Example 5.1.12 Using Properties of Summation and Products
Let 𝑎𝑘 = k + 1 and 𝑏𝑘 = k – 1 for every integer k.
Write each of the following expressions as a single summation or product:
a.
b.
Change of Variable
Observe that
and also that
Hence
More complicated changes of variable
Change of Variable
Example 5.1.13 Transforming a Sum by a Change of Variable
Transform the following summation by making the specified change of variable:
First calculate the lower and upper limits of the new summation:
when k  0, j  k  1  0  1  1.
when k  6, j  k  1  6  1  7.
Then the new sum goes from j = 1 to j = 7
1
1
1


k  1 (j  1 )  1 j
Finally,
6
1


k

1
k 0
7
1

j 1 j
Change of Variable
Example 5.1.14 When the upper limit appears in the expression to be summed
Rewrite the summation
so that the lower limit becomes 0 and the upper limit becomes n
but the index of the summation remains k.
a. First transform the summation by making the change of variable j = k – 1.
b. Second, transform the summation obtained in part (a) by changing all j’s to k’s
Change of Variable
Example 5.1.14 When the upper limit appears in the expression to be summed
Rewrite the summation
so that the lower limit becomes 0 and the upper limit becomes n
but the index of the summation remains k.
a. First transform the summation by making the change of variable j = k – 1.
When k = 1, then j = 1 – 1 = 0, and When k = n + 1, then j = n
So the new lower limit is 0 and the new upper limit is n.
Next,
Therefore,
b. Second, transform the summation obtained in part (a) by changing all j’s to k’s
Therefore,
Factorial and “n Choose r” Notation
Definition
For each positive integer n, the quantity n factorial denoted n!, is defined to be the
product of all the integers from 1 to n:
n! = n∙(n – 1) ∙ … ∙3 ∙2 ∙1
Zero factorial, denoted 0!, is defined to be 1:
0! = 1
Recursively,
Example 5.1.15 The First Ten Factorials
Factorial and “n Choose r” Notation
Example 5.1.16 Computing with Factorials
Simplify the following expressions:
Factorial and “n Choose r” Notation
Definition
Let n and r be integers with 0 ≤ r ≤ n. The symbol
n
 
r 
is read “n choose r” and represents the number of subsets of size r that can be chosen
from a set with n elements. (= nCr = Cn,r = C(n,r))
Formula for computing
n
 
r 
For integers n and r with 0 ≤ r ≤ n,
 n
n 
   n!  

r
!
(
n

r
)!
r 
n  r
Factorial and “n Choose r” Notation
Example 5.1.17 Computing
n
 
r 
Mathematical Induction
Principle of Mathematical Induction
Let P(n) be a property that is defined for integers n, and let a be a fixed integer.
Suppose the following two statements are true:
Basis step :
1. P(a) is true.
Inductive step :
2. For every integer k ≥ a, if P(k) is true then P(k + 1) is true.
Then the statement
for every integer n ≥ a, p(a)
is true.
Mathematical Induction
Method of Mathematical Induction
Consider a statement of the form,
“for all integer n ≥ a, a Property P(n) is true.”
To prove such a statement, perform the following two steps :
1. Basis Step
Show that the property P(n) is true for n = a.
2. Inductive Step Show that all integer k ≥ a,
if the property is true for n = k then it is true for n = k + 1.
To perform the inductive step,
Suppose that the property is true for n = k,
where k is any particular but arbitrarily chosen integer with k ≥ a.
Then show that the property is true for n = k + 1.
Mathematical Induction
Example 5.2.1 Sum of the First n Integers (Theorem 5.5.1)
Use mathematical induction to prove that
for every integer n ≥ 1.
Proof:
Basis Step:
P(1)  1 
1 (1  1)
2
Inductive Step:
Hypothesis : Suppose that P(k) is true, for some integer k with k ≥ 1. P(k )  1  2  ...  k 
Proof : Prove that P(k  1)  1  2  ...  k  (k  1) 
P(k  1)  1  2  ...  k  (k  1)
k (k  1)

 (k  1)
2
(k  1)( k  2)

2
(k  1)( k  2)
is ture.
2
k (k  1)
2
Mathematical Induction
Example 5.2.1 Sum of the First n Integers (Theorem 5.5.1)
Prove that
r n1  1
r 
,

r 1
i 0
n
i
for all integers n ≥ 0 and all real numbers r except 1.
Proof:
Basis Step: Show that the property is true for n = 0.
r 01  1
P(0)   r 
r 1
i 0
0
i
Inductive Step:
Hypothesis: Suppose the property is true for n = k.
Proof: Show that property
is true for n = k + 1.
Defining Sequences Recursively
A sequence can be defined in a variety of different warys
1. One informal way is to write the first few terms with the expectation
that the general pattern will be obvious.
3, 5, 7, 9, 11, 13, 15, …
2. A second way to define a sequence is to give an explicit formula for its nth term.
(1) n
an 
n 1
3. A third way to define a sequence is to use recursion
Defining Sequences Recursively
Definition
A recurrence relation for a sequence 𝑎0 , 𝑎1 , 𝑎2 , … is a formula that relates each term 𝑎𝑘 to certain of
its predecessors 𝑎𝑘 −1 , 𝑎𝑘 −2 , …, 𝑎𝑘 −𝑖 , where i is an integer with k – i ≥ 0. If i is a fixed integer,
the initial conditions for such a recurrence relation specify the values of 𝑎0 , 𝑎1 , 𝑎2 , …, 𝑎𝑖 −1 .
If i depends on k, the initial conditions specify the values of 𝑎0 , 𝑎1 , …, 𝑎𝑚 , where m is an integer with
m ≥ 0.
Fibonacci Numbers
an  an 1  an 2 (recurrence relation )
a0  1, a1  1
(initial conditions )
a2  a1  a0  1  1  2
a3  a2  a1  2  1  3
a4  a3  a2  3  2  5
Defining Sequences Recursively
Example 5.6.1 Computing Terms of a Recursively Defined Sequence
Define a sequence 𝑐0 , 𝑐1 , 𝑐2 , … recursively as follows:
For every integer k ≥ 2,
𝑐𝑘 = 𝑐𝑘 −1 + k𝑐𝑘 −2 + 1
recurrence relation
𝑐0 = 1 and 𝑐1 = 2
initial condition
Find 𝑐2 , 𝑐3 , and 𝑐4 .
𝑐2 = 𝑐1 + 2𝑐0 + 1 = 2 + 2∙2 + 1 = 5
𝑐3 = 𝑐2 + 3𝑐1 + 1 = 5 + 3 ∙2 + 1 = 12
𝑐4 = 𝑐3 + 3𝑐2 + 1 = 12 + 4 ∙5 + 1 = 33
Defining Sequences Recursively
Example 5.6.3 Show That a Sequence That Satisfy the same Recurrence Relation
Let 𝑎0 , 𝑎1 , 𝑎2 , … and 𝑏0 , 𝑏1 , 𝑏2 , … satisfy the recurrence relation that the kth term equals 3 times
the (k – 1)st term for every integer k ≥ 2:
1. 𝑎𝑘 = 3𝑎𝑘−1 and 𝑏𝑘 = 3𝑏𝑘 −1 .
But suppose that the initial conditions for the sequences are different:
2. 𝑎1 = 2 and 𝑏1 = 1.
Find (a) 𝑎2 , 𝑎3 , 𝑎4 and (b) 𝑏2 , 𝑏3 , 𝑏4 .
Defining Sequences Recursively
Example Showing that a Sequence Given by an Explicit Formula Satisfies a Certain Recurrence Relation.