chapter 4 sequences and mathematical induction. 4.1 sequences
TRANSCRIPT
Chapter 4
Sequences and Mathematical Induction
4.1
Sequences
Sequences
• The main mathematical structure used to study repeated processes is the sequence.
• The main mathematical tool used to verify conjectures about patterns governing the arrangement of terms in sequences is mathematical induction.
Example
• Ancestor counting with a sequence– two parents, four grandparents, eight great-
grandparents, etc.
– Number of ancestors can be represented as 2position
– Example: 23 = 8 (great grandparents), therefore parents removed three generations are great grandparents for which you have a total of 8.
Sequences
• Sequence is a set of elements written in a row as illustrated on prior slide. (NOTE: a sequence can be written differently)
• Each element of the sequence is a term.• Example– am, am+1, am+2, am+3, …, an
– terms a sub m, a sub m+1, a sub m+2, etc.– m is subscript of initial term– n is subscript of final term
Example
• Finding terms of a sequence given explicit formulas– ak = k/(k+1) for all integers k ≥ 1
– bi = (i-1)/i for all integers i ≥ 2a
– the sequences a and b have the same terms and hence, are identical
a1 = 1/(1+1) = ½ b2 = (2-1)/2 = ½
a2 = 2/(2+1) = 2/3 b3 = (3-1)/3 = 2/3
a3 = 3/(3+1) = 3/4 b4 = (4-1)/4 = 3/4
a4 = 4/5 b5 = 4/5
Example
• Alternating Sequence– cj = (-1)j for all integers j≥0
– sequence has bound values for the term.– term {-1, 1}∈
c0 = (-1)0 = 1
c1 = (-1)1 = -1
c2 = (-1)2 = 1
c3 = (-1)3 = -1
c4 = (-1)4 = 1…
Example
• Find an explicit formula to fit given initial terms– sequence = 1, -1/4, 1/9, -1/16, 1/25, -1/36, …– What can we observe about this sequence?
• alternate in sign• numerator is always 1• denominator is a square
– ak = ±1 / k2 (from the previous example we know how to create oscillating sign sequence, odd negative and even positive.
– ak = (-1)k+1 / k2
1/12 -1/22 1/32 -1/42 1/52 -1/62
a1 a2 a3 a4 a5 a6
Summation Notation
• Summation notation is used to create a compact form for summation sequences governed by a formula.
• the sequence is governed by k which has lower limit (1) and a upper limit of n.
• This sequence is finite because it is bounded on the lower and upper limits.
€
ak∑k=1
n
= a1 + a2 + a3 + a4 + ... + an
Example
• Computing summations
– a1 = -2, a2 = -1, a3 = 0, a4 = 1, and a5 =2.
€
ak∑k=1
5
= a1 + a2 + a3 + a4 + a5
€
ak∑k=1
5
= −2 + −1+ 0 +1+ 2 = 0
Example
• Computing summation from sum form.
€
k 2
k=1
5
∑
€
k 2
k=1
5
∑ =12 + 22 + 32 + 42 + 52 = 55
Example
• Changing from Summation Notation to Expanded form
€
(−1)i
i +1i=0
n
∑
€
(−1)i
i +1i=0
n
∑ =(−1)0
0 +1+
(−1)1
1+1+
(−1)2
2 +1+ ...+
(−1)n
n +1
Example
• Changing from expanded to summation form.• Find a close form for the following:
€
1
n+
2
n +1+
3
n + 2+ ...+
n +1
2n
€
1
n+
2
n +1+
3
n + 2+ ...+
n +1
n + n=
k +1
n + kk=0
n
∑
Separating Off a Final Term
• A final term can be removed from the summation form as follows.
• Example of use: Rewrite the following separating the final term
€
akk=m
n
∑ = akk=m
n−1
∑ + an
€
1
k 2k=1
n
∑ =1
k 2k=1
n−1
∑ +1
n2
Example
• Combining final term
€
2k + 2n
k=0
n−1
∑
€
2k + 2n
k=0
n−1
∑ = 2k
k=0
n
∑
Telescoping Sum
• Telescoping sum can be evaluated to a closed form.
€
1
k−
1
k +1=
(k +1) − k
k(k +1)=
1
k(k +1)
€
1
k(k +1)k=1
n
∑ =1
k−
1
k +1
⎛
⎝ ⎜
⎞
⎠ ⎟
k=1
n
∑ =1
1−
1
2
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
2−
1
3
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
3−
1
4
⎛
⎝ ⎜
⎞
⎠ ⎟+ ...+
1
n −1−
1
n
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
n−
1
n +1
⎛
⎝ ⎜
⎞
⎠ ⎟=1 −
1
n +1
Product Notation
€
akk=1
5
∏ = a1a2a3a4a5
€
akk=1
n
∏ = akk=1
n−1
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟ an Recursive form
Example
• Compute the following products:
€
kk=1
5
∏ =1*2* 3* 4 *5 =120
k
k +1k=1
1
∏ =1
1+1=
1
2
Factorial
• Factorial is 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 equal to 1.
Example
• Computing Factorials
€
8!
7!=
8* 7!
7!= 8
5!
2!*3!=
5* 4 * 3!
2!*3!=
5 * 4
2!
Properties of Summations and Products
• Theorem 4.1.1– If am, am+1, am+2, … and bm, bm+1, bm+2, … are
sequences of real numbers and c is any real number, then the following equations hold for any integer n≥m:
€
1. ak + bkk=m
n
∑k=m
n
∑ = ak +bk( )k=m
n
∑
2. c * akk=m
n
∑ = c * akk=m
n
∑
€
3. akk=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟* bk
k=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟= ak *bk( )
k=m
n
∏
Examples
• Let ak = k +1 and bk = k – 1 for all integers k
€
ak + 2⋅ bkk=m
n
∑k=m
n
∑
k +1+ 2⋅ k −1k=m
n
∑k=m
n
∑
k +1+ 2⋅ k −1( )k=m
n
∑
3k −1k=m
n
∑
Examples
• Let ak = k +1 and bk = k – 1 for all integers k
€
akk=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟• bk
k=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟= (k +1)
k=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟• (k −1)
k=m
n
∏ ⎛
⎝ ⎜
⎞
⎠ ⎟
(k +1)(k −1) = k 2 −1k=m
n
∏∏
Transforming a Sum by Change of Variable
• Transform the following by changing the variable.– summation: – change of variable: j = k+1– Solution:• compute the new limits:
– lower: j=k+1, j=0+1=1– upper: j=k+1, j=6+1=7
€
1
k +1k=0
6
∑
€
1
( j −1) +1=
j=1
7
∑ 1
jj=1
7
∑