Block 3 Discrete Systems Lesson 7a – Problem Exercises in

Block 3 Discrete Systems
Lesson 10 –Sequences and
Series
Both finite and countable infinite
series and much more
one two three four five six seven eight nine ten
Narrator: Charles Ebeling
University of Dayton
1
Summation Notation
n
a
im
 am  am 1  ...  an 1  an
i
n
 i  1  2  ...  n
i 1
n
 c  nc
i 1
n
 ca
i 1
i
n
 c  ai
i 1
n
n
n
a  b    a  b
im
i
i
i m
i
i m
i
2
Defining Sequences



Sn is the nth term in a sequence that may be finite or
infinite
Sn is a function defined on the set of natural numbers
Examples:
 If Sn = 1/n, then the first 4 terms in the sequence are
1, 1/2, 1/3, 1/4
 If Sn = 1/n!, then the first 4 terms in the sequence
are 1, 1/2, 1/6, 1/24
 The general term for the sequence -1, 4, -9, 16, -25 is
Sn = (-1)n n2
3
Arithmetic Progression


Sn = a + (n-1)(d) is an arithmetic progression
starting at a and incrementing by d
For example, the first six terms of Sn = 3 +
(n-1) (4) are 3, 7, 11, 15, 19, 23
4
The limit of an infinite sequence
If for an infinite sequence,
s1, s2, …, sn, …
there exists an arbitrarily small  > 0 and an
m > 0 such that |sn – s| <  for all n > m,
then s is the limit of the sequence.
lim sn  s
n 
5
Examples
2
1 2
n2  2
1
n
lim 2
 lim

n  2n  3n
n 
3 2
2
n
 2n  1 
2n  1
1 
1
lim 4  n1   4  lim n1  4  lim   n1   3.5
n 
n  2
n  2
2 
2 


6
Series


The sum of a sequence is called a series.
The sum of an infinite sequence is called an
infinite series


If the series has a finite sum, then the series is
said to converge; otherwise it diverges
A finite sum will always converge
Let Sn = s1 + s2 + … + sn
Sn is the sequence of partial sums
7
The Arithmetic Series
Sn  a  (a  d )  (a  2d )  (a  3d )  ... 
...   a  (n  3)d    a  (n  2)d   [a  (n  1)d ]
n
n
Sn   s1  sn   [2a  (n  1)d ]
2
2
The sum of the first 100 odd numbers is
100
Sn 
 2   99  2   10, 000
2
10
More to do with arithmetic series
n
n  n  1
i 1
2
i 
n  n  1 2n  1
i 

6
i 1
n
2
n  n  1
i 

4
i 1
2
n
3
2
For the overachieving student:
Prove these results by induction
11
The Geometric Sequence
2
a, ar , ar ,..., ar
n 1
examples
2, 2 .7  , 2 .7  , 2 .7  ,...2 .7  or 2, 1.4, 0.98, 0.686,...
2
3
n
2, 0.2, 0.02, 0.002; where a  2, r  0.1, n  4
12
The Geometric Series
Sn = a +ar + ar2 + …
+ arn-1
r Sn = ar + ar2 + … + arn-1 + arn
Sn - r Sn = (1-r) Sn = a - arn
Sn 
This is a
most
important
series.
a 1  r n 
1 r
a 1  r n 
a
lim S n 

, if r  1
n 
1 r
1 r
13
The Geometric Series in Action

Find the sum of the following series:
n
n

  1   1  2  1 3

1
1
10 1           ...     ...  10  
3
n 0  3 
  3   3   3 

10
30


 15
1  1/ 3 2

a
n
ar 
, if r  1

1 r
n 0
14
Future Value of an Annuity
The are n annual payments of R (dollars) where the
annual interest rate is r.
Let S = the future sum after n payments, then
S = R + R(1+r) + R(1+r)2 + … + R(1+r)n-1
n 1
n 1
i 0
i 0
S   (1  r )i R  R  (1  r )i
0
1
2
…
n-2
n-1
n
0
R
R
…
R
R
R
R(1+r)
R(1+r)2
R(1+r)n-1
15
More Future Value of an Annuity
n
n
1

(1

r
)
(1

r
)
1
i
i
S   (1  r ) R  R  (1  r )  R
R
1  1  r 
r
i 0
i 0
n 1
n 1
the sum of a finite geometric series
s  a  ar  ar 2  ...  ar n 1 
a 1  r n 
1 r
; r 1
16
The Binomial Theorem
A really Big Bonus. Isaac
Newton’s first great discovery
(1676)
17
First, a notational diversion…
Factorial notation: n! = 1·2·3 ··· (n-2) ·(n-1) ·n
where 0! = 1! = 1 and n! = n (n-1)!
 n  n(n  1)(n  2)  (n  r  1)
n!

 
1 2  3  (r  1)  r
r !(n  r )!
r 
for example:
8
8  7  6  5  4  3  2 1
8!
87 6


 56
 
 3  1 2  3 5  4  3  2 1 3!(5)! 1  2  3
A useful fact:
n  n 
n!


  

r
n

r
  
 r !(n  r )!
18
Here it is…For n integer:
n(n  1) n  2 2 n(n  1)(n  2) n 3 3
(a  b)  a  na b 
a b 
a b
1 2
1 2  3
n(n  1) 2 n  2
 ... 
a b  nab n 1  b n
1 2
n
 n  nr r
   a b
r 0  r 
n
n
n 1
 6 6  6 5  6
 6 3 3  6 2 4  6 5  6 6
2
 a  b     a    a b    a 4b    a b    a b    ab    b
0
1 
 2
3
 4
5
 6
 a 6  6a5b  15a 4b 2  20a 3b3  15a 2b 4  6ab5  b6
6
19
Some Observations on the
binomial theorem




n+1 terms
sum of the exponents in each term is n
coefficients equi-distant from ends are equal
A related theorem:
 n  1  n   n 


 
 r   r  1  r 
This is truly an
amazing result! Look
for my triangle on the
next slide.
20
Pascal’s triangle
 n  1  n   n 


 
 r   r  1  r 
5  4  4
     
3  2 3 
10  6  4
21
The Generalized Binomial Theorem
For n non-integer or negative:
 n  nr r
( a  b)     a b
r 0  r 

n
22
Other series worth knowing
(1  x) 1  1 x  x 2
x3  x 4
x 5  ...
x
2
 1 binomial series
2
3
n

x
x
x
e x  1  x    ...  
2! 3!
n 0 n !
x 2 x3 x 4
ln 1  x   x     ...  1  x  1
2 3 5
exponential series
logarithmic series
23
This Series has come to an
end
Next Time – Discrete Probability
24