415 lecture note #7 

[Ch 5] Recurrence Relations (1)

5.1 Intro

• Consider the following sequence of numbers:

  a₁= 5, a₂= 8, a₃= 11, a₄= 14, a₅= 17, ...

  It seems there is a relation between any two adjacent terms a_n and a_n+1, namely a_n+1= a_n+ 3, for any n (where n >= 1).

  There are two ways to express the relation above:

  1. Recursively

     

  2. Directly

     

• A recurrence relation is defined by expressing the n^th term (a_n) in terms of its predecessor(s) (a_n-1).  A recurrence relation also includes initial conditions (as base case(s)) where the sequence starts.

   

• Example 1:  Compound interest

  With an initial amount M and r% annual interest, the amount at the end of the nth year, A_n, is expressed recursively as:

  A_n = (1 + r)*A_n-1, for all n >= 1

  with initial condition

  A₀ = M

• Example 2:  Fibonacci numbers

  f_n = f_n-1 + f_n-2, for all n >= 3

  with initial conditions

  f₁ = 1 and f₂ = 2

   

5.2 Solving Recurrence Relations

• Closed form of a recurrence relation
  • We are also interested in knowing a direct form, since the recurrence relation does not show it explicitly.
  • Direct forms do not involve recurrence -- closed forms.
  • Closed forms allows us to plug in a value for n and get the result immediately. 

  Example:  Sequence 5, 8, 11, 14, 17, ...

  Closed form is a_n = 3n + 2, for all n >= 1.  So,

  a₁ = 3*1 + 2 = 5
  a₂ = 3*2 + 2 = 8  etc.
  

• There are two ways to derive a closed form for a given recurrence relation.
  1. Iteration
  2. Characteristic functions

(1) Iteration Method

• Steps:
  1. Write the recurrence equation for a_n (This equation has terms in a_n-1, a_n-2, etc.)
  2. Replace each of a_n-1, a_n-2 by its recursive expression
  3. Continue until you see a pattern developing
  4. Prove (usually by induction) that the pattern found is correct.

      

• Example 1:

  a_n = a_n-1 + 3, for all n >= 2

  with initial condition a₁ = 5

  a_n = a_n-1 + 3
               [a_n-1 = a_n-2 + 3 ... as a side note
      = a_n-2 + 3 + 3         ... substitute a_n-2 + 3 for a_n-1
      = a_n-3 + 3 + 3 + 3   ... substitute a_n-3 + 3 for a_n-2
      ...
      = a_n-k + k*3            ... derive a pattern for an arbitrary kth iteration
     ...
      = a₁ + (n-1)*3         ... k = (n-1) since n-(n-1) = 1

  Then, we can plug in a₁ = 5.  We get

  a_n = 5 + (n-1)*3 = 3n - 3 + 5 = 3n + 2  ... closed form

   

• Example 2:

  S_n = 2*S_n-1, for all n >= 1

  with initial condition S₀ = 1

  S_n = 2*S_n-1
               [S_n-1 = 2*S_n-2 ... side note
       =

   

   

   

   

   

• Example 3:   Tower of Hanoi

  Number of moves required for n disks is

  C_n = 2C_n-1+ 1, for all n >= 2

  with initial condition C₁ = 1.

  C_n = 2C_n-1+ 1
               [C_n-1 = 2C_n-2 + 1 ... side note
        = 2(2C_n-2 + 1) + 1           ... substitute 2C_n-2 + 1 for C_n-1
        = 4C_n-2 + 2 + 1
        = 4(2C_n-3 + 1) + 2 + 1     ... substitute 2C_n-3 + 1 for C_n-2
        = 8C_n-3 + 4 + 2 + 1
        ....
        = 2^kC_n-k + 2^k-1 + ...+ 2¹ + 2⁰                ... a general pattern!!
        ....
        = 2^(n-1)*C₁ + 2^n-2 +2^n-3 + ...+ 2¹ + 2⁰   ... k = (n-1) since n-(n-1) = 1

  Then we can plug in C₁ = 1 in the above, obtaining

  C_n = 2^n-1 + 2^n-2 +2^n-3 + ...+ 2¹ + 2⁰
            2ⁿ - 1
        = --------       ... by geometric summation formula
              2 - 1
        = 2ⁿ - 1

   

• More examples
  1. T(n) = -3*T(n-1), T(0) = 2
  2. T(n) = T(n-1) + 1, T(0) = 0
  3. T(n) = 2*T(n-1) + 3, T(0) = 0
  4. T(n) = T(n/2) + 1, T(1) = 0