Infinite Sequences
Key Questions
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It depends on the type of sequence.
If the sequence is an arithmetic progression with first term
#a_1# , then the terms will be of the form:#a_n = a_1 + (n-1)b#
for some constant b.If the sequence is a geometric progression with first term
#a_1# , then the terms will be of the form:#a_n = a_1 * r^(n-1)#
for some constant#r# .There are also sequences where the next number is defined iteratively in terms of the previous 2 or more terms. An example of this would be the Fibonacci sequence:
#0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...# Each term is the sum of the two previous terms.
The ratio of successive pairs of terms tends towards the golden ratio
#phi = 1/2 + sqrt(5)/2 ~= 1.618034# The terms of the Fibonacci sequence are expressible by the formula:
#F_n = (phi^n-(-phi)^-n)/sqrt(5)# (starting with#F_0 = 0# ,#F_1 = 1# )In general an infinite sequence is any mapping from
#NN -> S# for any set#S# . It can be defined in any way you like.Finite sequences are the same, except that they are mappings from a finite subset of
#NN# consisting of those numbers less than some fixed limit, e.g.#{n in NN: n <= 10}# -
A sequence is an infinite set of points which are ordered and a mapping can be associated with with that set and the set of natural numbers.
The general notion of a sequence is that it is an infinite set with every element associated with a natural number even though all infinite sets may not be sequences.
A sequence may be represented by,
#{x_n}# where#x_n# is the#n# th element related to the a corresponding natural number.Thus if
#x_n = 1/n^2# , the sequence may be given as,#{1,1/4,1/9,1/16,....}# For
#x_n = n^3# we shall have,#{1, 8, 27, 64,....}# Now for
#x_n = n# we can have,#{1,2,3,.....}# This is indeed the set of naturals.
However, there can be other ordered arrays of numbers which are sometimes referred at as sequences. They don't fit right with the definition I gave.
Let's for example take a Fibonacci sequence.
#1,1,2,3,5,8,13,....# This sequence is made by adding the previous two numbers on the list to form the next one and so on.
There can be arithmetic sequences, like
#2,8,14,20,....# which has first term#2# and common difference#6# .I like to call them progressions and reserve the word sequence for the definition I proposed in the first 2 paragraphs.
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Answer:
It depends.
Explanation:
There are many types of sequences. Some of the interesting ones can be found at the online encyclopedia of integer sequences at https://oeis.org/
Let's look at some simple types:
#a_n = a_0 + dn# e.g.
#2, 4, 6, 8,...# There is a common difference between each pair of terms.
If you find a common difference between each pair of terms, then you can determine
#a_0# and#d# , then use the general formula for arithmetic sequences.#a_n = a_0 * r^n# e.g.
#2, 4, 8, 16,...# There is a common ratio between each pair of terms.
If you find a common ratio between pairs of terms, then you have a geometric sequence and you should be able to determine
#a_0# and#r# so that you can use the general formula for terms of a geometric sequence.Iterative Sequences
After the initial term or two, the following terms are defined in terms of the preceding ones.
e.g. Fibonacci
#a_0 = 0#
#a_1 = 1#
#a_(n+2) = a_n + a_(n+1)# For this sequence we find:
#a_n = (phi^n - (-phi)^(-n))/sqrt(5)# where#phi = (1+sqrt(5))/2# There are many ways to make these iterative rules, so there is no universal method to provide an expression for
#a_n# Polynomial Sequences
If the terms of a sequence are given by a polynomial, then given the first few terms of the sequence you can find the polynomial.
e.g.
#color(red)(1), 2, 4, 7, 11,...# Form the sequence of differences of these values:
#color(red)(1), 2, 3, 4,...# Form the sequence of differences of these values:
#color(red)(1), 1, 1,...# Once you reach a constant sequence like this, pick out the initial terms from each sequence. In this case
#1# ,#1# and#1# .These form the coefficients of a polynomial expression:
#a_n = color(red)(1)/(0!) + (color(red)(1)*n)/(1!) + (color(red)(1)*n(n-1))/(2!)# #=n^2/2+n/2+1#