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Sequences

  • A sequence is a function from a subset of the set of integers (usually either the set {0,1,2,...} or the set {1,2,3,...}) to a set S
  • We use the notation an to denote the image of the integer n
  • We call an a term of the sequence (term = phần tử)
  • We use the notation {an} to describe the sequence

Geometric progression (Cấp số nhân)

  • A geometric progression is a sequence of the form
a,ar,ar2,...,arn,...
  • where the initial term a and the common ratio r are real numbers.
  • In other words, A geometric progression is a discrete analogue of the exponential function
f(x)=arx
  • Example:
    • The sequences {dn} with dn=6(13)n
      • a = 6
      • r = 1/3

Arithmetic progression (Cấp số cộng)

  • An arithmetic progression is a sequence of the form
a,a+d,a+2d,...,a+nd,...
  • where the initial term a and the common difference d are real numbers.
  • In other words, A arithmetic progression is a discrete analogue of the linear function
f(x)=dx+a
  • Example:
    • The sequences {sn} with sn=1+4n
      • a = -1
      • d = 4

Recurrence Relations

  • A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence
  • a0,a1,...,an1, for all integers n with nn0, where n0 is a nonnegative integer
  • A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation
  • Example:
    • Let {an} be a sequence that satisfies the recurrence relation an=an1an2 for n=2,3,4,... , and suppose that a0=3 and a1=5
    • a2=2, a3=3
  • The a0 and a1 from the example is called initial conditions

Fibonacci sequence

  • Initial conditions f0=0,f1=1
  • Recurrence relation
fn=fn1+fn2 for n=2,3,4,...

Closed formula

  • We say that we have solved the recurrence relation together with the initial conditions when we find an explicit formula
  • Example:
    • Let {an} be a sequence that satisfies the recurrence relation an=nan1 for n=1,2,3,4,... , and suppose that a1=1
    • an=n!, the factorial function

Solve recurrence relations

Iteration

  • Let {an} be a sequence that satisfies the recurrence relation an=an1+3 for n=1,2,3,4,... , and suppose that a0=2

Forward substitution

a1=2+3a2=(2+3)+3=2+32a3=(2+23)+3=2+33an=2+3(n1)

Backward substitution

an=an1+3=(an2+3)+3=an2+3.2=(an3+3)+3.2=an3+3.3=2+3(n1)

Special Integer Sequences

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  • Example:
    • How can we produce the terms of a sequence if the first 10 terms are 1,3,4,7,11,18,29,47,76,123 ?
    • an=an1+an2 with initial conditions are a0=1, a1=3

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