Sequences and Series - some ideas with Matlab

For example, the sequence of odd integers is generated by the sequence 
2n – 1. The sequence produce 2(1)-1, 2(2)-1, 2(3)–1, 2(4)-1... which produces 1, 3, 5, 7... 

A series is the sum of the terms of a sequence. Series are based on sums, whereas sequences are not. 

In Matlab, let’s generate the sequence above (for odd numbers), in two ways:
 

a) Using a for-loop in a sequence:

% We initialize our sum
s = 0;
% We are going to get the first 5 terms,
% one-by-one in our sequence
for n = 1 : 5
    % This is the shown formula for odd numbers
    tn = 2*n-1
    % We add-up every term in the series
    s = s + tn;
end

% We display the sum of the series
s

The code produces these results: 

tn = 1
tn = 3
tn = 5
tn = 7
tn = 9
s =  25

b) Using a vectorized way, which is simpler and faster:

% We define our first 5 ns
n = 1 : 5
% We calculate the terms
tn = 2*n - 1
% We add-up all the terms in the series
s = sum(tn)
 

The code produces these results:

n =   1     2     3     4     5
tn =  1     3     5     7     9
s =   25

Arithmetic Sequences

This is a sequence of numbers each of which (after the first) is obtained by adding to the preceding number a constant called the common difference. Then, 3, 6, 9, 12, 15... is an arithmetic sequence because each term is obtained by adding 3 to the preceding number. In the arithmetic sequence 500, 450, 400... the common difference is -50.

Formulas for arithmetic sequences:
- The n^th term, or last term: L = a + (n - 1)d
- The sum of the first n terms: S = n/2 (a + L) = n/2 [2a + (n - 1)d]

where
a = first term of the sequence
d = common difference
n = number of terms
L = n^th term, or last term
S = sum of first n terms
 

Example:

Consider the arithmetic sequence 3, 7, 11, 15...
where a = 3 and d = 4.

The sixth term is:

L = a + (n - 1)d = 3 + (6 - 1)4 = 23.

The sum of the first six terms is:

S = n/2(a + L) = 6/2(3 + 23) = 78 or
S = 6/2[2a + (n - 1)d] = 6/2[2(3) + (6 - 1)4] = 78.

Let’s do it effortlessly with Matlab, without for-loops (vectorized way):
 

% Let's define our sequence, starting with 3,
% using steps of four units, until we reach 999, for example
seq = 3 : 4 : 999;

% Find the 6th element
sn = seq(6)

% Find the sum of the first 6 elements
s1_6 = sum(seq(1 : 6))
 

The results are: 

sn = 23
s1_6 =  78 

Geometric Sequences

A geometric sequence is a sequence of numbers each of which is obtained by multiplying the preceding number by a constant number called the common ratio. So 4, 8, 16, 32... is a geometric sequence because each number is obtained by multiplying the preceding number by 2. In the geometric sequence 64, 16, 4, l, 1/4... the common ratio is 4.

Formulas for geometric sequences:
- The n^th term, or last term: L = ar^n-1
- The sum of the first n terms:

S = a(rⁿ – 1)/(r - 1) = (rL-a)/(r - 1)  (r and 1 are different)

where
a = first term
r = common ratio
n = number of terms
L = nth term, or last term
S = sum of first n terms
 

Example:

Consider the geometric sequence 5, 10, 20, 40...
where a = 5 and r = 2 

The seventh term is:
L = ar^n-1 = 5(2^7-1) = 5(2⁶) = 320

The sum of the first seven terms is:
S = a(r^n-1)/(r-1) = 5(2⁷-1)/(2-1) = 635

 
Let’s calculate it with Matlab:

% Define your important constants
a = 5; r = 2;

% Define your sequence
n = 1 : 7;
L = a * r.^(n-1)

% Find the 7th term
L(7)

% Find sum of first 7 terms
sum(L(1:7))

Matlab response is:

L =      5    10    20    40    80   160   320
ans =    320
ans =    635
 

Another way to approach it is: 

% Define your exponents
n = 0 : 6

% Define your sequence
gs = 2.^n * 5

% Find the 7th term
gs(7)

% Find the sum of the first 7 terms
sum(gs(1 : 7))

The answer is: 

n =      0     1     2     3     4     5     6
gs =     5    10    20    40    80   160   320
ans =    320
ans =    635