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Infinite Sequences and Summation Notation
Definition: A sequence is a function whose domain is the set of natural numbers Z+
Notation is generally like this: {an} where a1
Ex 1 Sketch the graph of the sequence
= f(1), a2 = f(2), a3 = f(3), ..., an = f(n), ...
1
{ }
n
Ex 2 List the first three terms and the 100th term of the sequence:
A)
{
n
n+1
}
B) { 2 + ( .1)
C) { ( −1 )
{
n
n+1
n
}
n+ 1
n
2
n+1
}=
}
1 2 3
100
, , , ..
2 3 4
101
n
{ 2 + ( .1) } = 2.1 , 2.01 , 2.001 , .... , 2.000000000000000000000...1
{ ( −1)
n+ 1 n
2
n+1
}=
4 9
−10000
, − , , .... ,
2 3 4
101
1
Definition recursion is when a function will use itself as part of its definition.
In the computer field, it's when a function 'calls' itself.
An example of this is the 'factorial' function n! since n! = n ⋅ ( n − 1 )!. Here
n factorial is defined as the product of n and n-1 factorial.
Another common use of recursion is the 'Fibonacci' numbers. Here the
'next' number is the sum of the two that preceded it: a
=a
+ a but
n+ 2
n+ 1
n
start with a = a = 1
1
2
(Did anyone read the book 'The DaVinci Code'?)
Ex 3 List the first 12 Fibonacci numbers.
1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , ..
Ex 4 If we assume that as n -> ∞ the ratio between terms of the Fibonacci numbers
approaches a constant, find that constant. (Note it is called the golden mean of
architecture and there are many real world places where this number occurs.)
Let the constant ratio equal φ. Then a
n+ 2
=a
2
n+ 1
+ a can be written as φ ⋅ a = φ ⋅ a + a
n
n
n
(Can you see why that is true?)
2
φ ⋅a = φ ⋅a + a
n
n
n
2
φ ⋅a − φ ⋅a − a = 0
n
(
n
n
)
2
a ⋅ φ −φ −1 =0
n
φ =
−( −1 ) +
2
( −1 ) − 4 ⋅ 1 ⋅ ( −1 )
2⋅ 1
=
1+
5
2
Ex 5 list the first 5 terms of the recursively defined sequence: a = 2 , a
1
k+ 1
= 3a
k
a = 3 ⋅ a = 3⋅ 2 = 6
2
1
a = 3 ⋅ a = 3 ⋅ 6 = 18
3
2
a = 3 ⋅ a = 3 ⋅ 18 = 54
4
3
a = 3 ⋅ a = 3 ⋅ 54 = 162
5
4
2 , 6 , 18 , 54 , 162 , ..
Ex 6 write the first 5 terms and the nth term of the recursively defined sequence:
a =1 a
= (n + 1)⋅ a
1
n+ 1
n
n
a = 2 ⋅ a = 2⋅ 1 = 2
2
1
a = 3 ⋅ a = 3⋅ 2 = 6
3
2
a = 4 ⋅ a = 4 ⋅ 6 = 24
4
3
a = 5 ⋅ a = 5 ⋅ 24 = 120
5
a
4
n+ 1
= ( n + 1)⋅ a = (n + 1)⋅ n⋅ a
n
n− 1
= ( n + 1 )n ( n − 1 ) ⋅ a
n− 2
= .... = ( n + 1 ) ( n − 1 ) ( n − 2 ) ....⋅ 2 ⋅ 1 = ( n +
m
Summation notation
∑
a = a + a + .... + a
k
1
2
m
k =1
6
Ex 7 Evaluate the sum:
∑
2
j ( j − 2)
j =2
6
∑
2
2
2
2
2
2
j ( j − 2 ) = 2 ⋅ ( 2 − 2) + 3 ⋅ ( 3 − 2) + 4 ( 4 − 2) + 5 ( 5 − 2) + 6 ( 6 − 2)
j=2
6
∑
2
j ( j − 2 ) = 0 + 9 + 32 + 75 + 144 = 260
j=2
5
Ex 8 Evaluate the sum
∑
( k + 2) ( k − 3 )
k =1
5
∑
( k + 2 ) ( k − 3 ) = 3 ⋅ ( −2 ) + 4 ( −1 ) + 5 ⋅ 0 + 6 ⋅ 1 + 7 ⋅ 2 = 10
k =1
5
Note that
∑
( j + 2 ) ( j − 3 ) yields the exact same sum as Ex 8. The change in
j=1
subscripting variable does not affect the sum.
Definition For the sequence { a } let us notate the following sequence:
n
S =a
1
1
S =a + a =S + a
2
2
1
1
2
S =a + a + a =S + a
3
3
2
1
2
3
.
.
n
S =S
n− 1
n
∑
+ a = a + a + .... + a =
n
1
2
n
a
j
j =1
Now the number S is called the nth partial sum of { a } and the sequence
n
n
{Sn } is called the sequence of partial sums of { a }.
n
Ex 9 Find a formula for the sum of the first n integers and use it to find the sum of the first
100 positive integers
S = 1 , S = 1 + 2 = 3 , S = ..
1
2
3
S = 1 + 2 + 3 + .... + ( n − 2 ) + ( n − 1 ) + n
n
S = n + ( n − 1 ) + ( n − 2 ) + .... + 3 + 2 + 1
n
2 S = ( n + 1) ⋅ n
n
S =
(n + 1)⋅ n
n
S
100
2
( 100 + 1 ) ( 100 )
=
2
= 5050
Theorem on the sum of a constant:
n
∑
(A)
n
C = n⋅ C
(B)
k =1
∑
C = (n − m + 1)⋅ C
k =m
Proof:
n
(A)
∑
C = C + C + .... + C = n ⋅ C
k =1
n
(B)
∑
k =m
m−1
n
C=
∑
k =1
C−
∑
k =1
C = n⋅ C − (m − 1)⋅ C = (n − m + 1)⋅ C
Note that summation notation may be used to denote polynomials:
n
∑
j
2
3
n
a ⋅ x = a + a ⋅ x + a ⋅ x + a ⋅ x + .... + a ⋅ x
j
0
1
2
3
n
j =0
Theorem on sums: If a , a , .... , a , .... and b , b , .... , b , .... are infinite
1
2
n
1
2
n
sequences and c is any real number, then for each positive integer n:
n
(A)
n
∑ (a j + b j) = ∑
a +
∑ (a j − b j) = ∑
a −
j =1
n
(B)
∑
b
j
j =1
n
j
j=1
n
∑
j
j =1
n
j =1
(C)
n
∑
b
j
j=1
n
c⋅ a = c⋅
j
j =1
∑
a
(proofs all direct and easy)
j
j =1
Theorem on sums: (alternate form) If a , a , .... , a , .... and b , b , .... , b , .... are
1
2
n
1
2
n
infinite sequences and c is any real number, then for each positive integer n:
n
∑(
c⋅ a + b
j
j=1
)
j
n
= c⋅
∑
n
a +
j
j=1
Ex 10 The equation 2x +
∑
b
j
(this covers all three in the above theorem)
j=1
1
= 0 has a root near 0.Find a procedure to
4
x + x+ 2
approximate the solution and use it to solve the equation with accuracy of 10-4.
1
2x = −
4
x + x+ 2
x=−
(
1
4
2⋅ x + x + 2
)
Let x = 0 and set up the procedure x = −
1
n
1
(
2⋅  x
)
n− 1
4
+x
n− 1
+ 2

x =−
2
x =−
3
x =−
4
1
(
2⋅  x
(
2⋅  x
5
x =−
6
x =−
8
2
)
4
+ x + 2
3
)
4
+ x + 2
3
=−

1
2 ⋅ ( x ) + x + 2
4
 4

1
2 ⋅ ( x ) + x + 2
5
 5

4
1
2 ⋅ ( x ) + x + 2
6
 6

4
=−

2
4
x =−
=−
2 ⋅ ( 0 ) + 0 + 2
1
x =−
7
1
4
= −.25
1
 1 4

1
2 ⋅  −  + − + 2
4
 4 

= −.2850779510
2 ⋅ ( −.2850779510) + −.2850779510 + 2
=−
=−
2 ⋅ ( x ) + x + 2
7
 7

4
1
=−
1
4
1
4
1
2 ⋅ ( −.2904398634) + −.2904398634 + 2
4
1
2 ⋅ ( −.2912605598) + −.2912605598 + 2
4
1
2 ⋅ ( −.2913861538) + −.2913861538 + 2
=−
4
= −.2912605598
= −.2913861538
= −.2914053732
1
2 ⋅ ( −.2914053732) + −.2914053732 + 2
4
The solution (good to 4 places) is x = -.2914
Check of the answer using software:
2x = −
= −.2904398634
1
4
x + x+ 2
 −.74679490189114522028 − .63092526448515280527⋅ i 
 −.74679490189114522028 + .63092526448515280527⋅ i 


−.29140884578315660891


 .89249932478272352474 − .99932504149695877662⋅ i 


 .89249932478272352474 + .99932504149695877662⋅ i 
= −.2914083144
Ex 11 The sequence defined recursively by a
k+ 1
(
= c⋅ a 1 − a
k
k
) is used in the study of
insect population growth. The constant c is called the Malthusian factor. Suppose that
1000⋅ a is the number of insects after k time intervals and that initially a = .25 measured
k
1
in thousands. Describe the behavior of the insect population for each value of c:
(A) c = .5
(A)
(B) c = 1.5
( 1) = .5⋅ .25⋅ (1 − .25) = .09375
a = .5 ⋅ a ⋅ ( 1 − a ) = .5 ⋅ .09375⋅ ( 1 − .09375) = .04248046875
3
2
2
a = .5 ⋅ a ⋅ ( 1 − a ) = .5 ⋅ .04248046875⋅ ( 1 − .04248046875) = .020337939262390136719
4
3
3
a = .5 ⋅ a ⋅ ( 1 − a ) = .5 ⋅ .020337939262390136719⋅ ( 1 − .020337939262390136719) = .009962153743
5
4
4
a = .5 ⋅ a ⋅ 1 − a
2
1
(
a = .5 ⋅ a ⋅ 1. − a
6
5
(
) = .5⋅ .009962153743⋅ (1. − .009962153743) = .004.931454619
5
a = .5 ⋅ a ⋅ 1. − a
7
6
) = [.5⋅ .004931454619⋅ ( 1. − .004931454619) ] = .002453567688
6
This number keeps getting smaller, so in time the population will be extinct.
(B)
(
1
) = 1.5⋅ .25⋅ (1 − .25) = .28125
(
2
) = 1.5⋅ .28125⋅ (1 − .28125) = .3032226563
(
3
) = 1.5⋅ .3032226563⋅ (1 − .3032226563) = .3169180155
(
4
) = 1.5⋅ .3169180155⋅ (1 − .3169180155) = .3247214805
(
5
) = 1.5⋅ .3247214805⋅ (1 − .3247214805) = .3289161609
a = 1.5⋅ a ⋅ 1 − a
2
1
a = 1.5⋅ a ⋅ 1 − a
3
2
a = 1.5⋅ a ⋅ 1 − a
4
3
a = 1.5⋅ a ⋅ 1 − a
5
4
a = 1.5⋅ a ⋅ 1 − a
6
5
(
6
) = 1.5⋅ .3289161609⋅ (1 − .3289161609) = .3310954800
(
7
) = 1.5⋅ .3310954800⋅ (1 − .3310954800) = .3322068947
(
8
) = 1.5⋅ .3322068947⋅ (1 − .3322068947) = .3327682108
a = 1.5⋅ a ⋅ 1 − a
7
6
a = 1.5⋅ a ⋅ 1 − a
8
7
a = 1.5⋅ a ⋅ 1 − a
9
a
a
a
a
a
8
(
10
= 1.5⋅ a ⋅ 1 − a
11
= 1.5⋅ a ⋅ 1 − a
12
= 1.5⋅ a ⋅ 1 − a
13
= 1.5⋅ a ⋅ 1 − a
14
= 1.5⋅ a ⋅ 1 − a
9
10
11
12
13
) = 1.5⋅ .3327682108⋅ (1 − .3327682108) = .3330502930
9
(
10
) = 1.5⋅ .3330502930⋅ (1 − .3330502930) = .3331916930
(
11
(
12
(
13
) = 1.5⋅ .3331916930⋅ (1 − .3331916930) = .3332624831
) = 1.5⋅ .3332624831⋅ (1 − .3332624831) = .3332979007
) = 1.5⋅ .3332979007⋅ (1 − .3332979007) = .3333156152
Enough yet? Looks like the population will be steady at about 333.
1)!