Chapter Two Notes
Basic Structures::
Set, Functions, Sequences, Sums and Matrices
Based on: Discrete Math & Its Applications - Kenneth Rosen
CSC125 - Spring 2012
2.4 Sequences and Summations
Definitions:
Sequence
A function from a subset of Z to a set S.
Often that subset is {0, 1, 2, . . } or {1, 2, 3, . . }
We usually denote the elements of S as an.
I.e., f(n) = an.
And we use {an} to denote the sequence.
Example: The sequence {an}, where an = 1/n is: 1, 1/2, 1/3, 1/4, . .
Geometric progression
a, ar, ar2, ar3, . . , arn,. .
a – initial term
r – common ratio
Note: r = an / an-1 = arn/ arn-1
Example: The sequence {an}, where an = 2*3n
Initial term is: 2
r=3
The terms a0, a1, a2, a3, . . are: 2, 6, 18, 54, . .
Arithmetic progression
a, a+d, a+2d, a+3d, . . , a+nd, . .
a – initial term
d – common difference
Note: d = an – an-1 = (a+nd) – (a+(n-1)d)
1
Example: The sequence {an}, where an = –5 + 3n
Initial term is: –5
d=3
The terms a0, a1, a2, a3, . . are: –5, –2, 1, 4, . .
Summation
Notation:: for the sum of the terms am, am+1, . ., an, from the
sequence {an}, we can write:
n j=m aj or  m ≤ j ≤ n aj or as in Rosen, p. 163.
index of summation
In the above example, j is the index of summation.
We could have used other letters, such as i or k, as below:
n i=m ai or  m ≤ i ≤ n ai or n k=m ak or  m ≤ k ≤ n ak.
Examples:
5 i=1 i = 1 + 2 + 3 + 4 + 5 = 15
5 k=2 1/k = 1/2 + 1/3 1/4 + 1/5 = (30+20+15+12)/60 = 77/60
5 k=0 3 = 3 + 3 + 3 + 3 + 3 + 3 = 18 {Weird, but legitmate!}
3 j=0 2*3j = 2+6+18+54 = 80 = (162 – 2)/2 {Table 2, p. 166}
10 k=3 –5 + 3k = 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 = 116
= 8 + 3*(8*9/2), because
4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 =
(1+3) + (1+2*3) + (1+3*3) + . . (1+8*3) =
8 + 3 + 2*3 + 3*3 + . . + 8*3 = 8 + 3(1+2+3+ . . +8) =
8 + 8 k=1 k = 8 + (8*9/2) {Table 2, p. 166}
Double summation
Perform inner summation to obtain a sequence,
then perform the outer summation on the sum of that
sequence.
Examples:
2
4 i=1 3 j=0 i*j = 4 i=1 (0*i + 1*i + 2*i + 3*i) =
4 i=1 6i = 6 + 12 + 18 + 24 = 60
Note: Since 6i represents an arithmetic progression, we can
factor out 6 and use the formula for 4 i=1 i
6*(5*4/2) = 60
4 i=1 3 j=0 i – j = 4 i=1 (i + i – 1 + i – 2 + i – 3) =
4 i=1 4i – 6 = (4 – 6) +(8 – 6) +(12 – 6) +(16 – 6) = 16
4 i=1 3 j=0 i = 4 i=1 (i + i + i + i) = 4 i=1 4i =
4 + 8 + 12 + 16 = 40
4 i=1 3 j=0 j = 4 i=1 (0 + 1 + 2 + 3) = 4 i=1 6 =
6 + 6 + 6 + 6 = 24
Note: Study Tables 1 & 2, pp. 162 & 166
Theorem 1, p. 164
Cardinality
countable
uncountable
3