Linear Recurrence Relations Part II

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Linear Recurrence RelationsPart II

• Vasileios Hatzivassiloglou
• University of Texas at Dallas

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Review Problems

• Suppose that you have unlimited pegs for the
  Towers of Hanoi puzzle. How many moves are needed
  for solving it for n disks?
• What is the characteristic equation for the
  recurrence relation an 3an-1 2an-3?
• Solve the recurrence relation an 8an-1 15an-2
  with the initial conditions a0 0, a1 2.

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The Fibonacci sequence

• fn fn-1 fn-2, f00 and f11
• Characteristic equation r2 r 1 0
• with roots
• and coefficients
• Finally,

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Fibonacci as a recursion example

• Often used in elementary programming texts to
  introduce the concept of recursion
• Evaluating fn via the recursive implementation
  takes
• exponential time and space!
• a very inefficient way to compute it
• Evaluating the closed form takes
• constant time

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Degenerate characteristic equation

• What happens if r1r2 (one root of multiplicity
  two)?
• Theorem 2 If the characteristic equation of a
  second degree linear homogeneous recurrence
  relation has only one root r0, then all solutions
  are of the form anb1r0nb2nr0n for n0, where b1
  and b2 are constants

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Example

• an 6an-1 9an-2, a01 and a16
• Characteristic equation r2 6r 9 0
• with only one root
• We again solve from the initial conditions
• a0 1 b1r00b20r00 b1
• a1 6 b1r01b21r01 133b2
• therefore b2 (6-3)/3 1
• giving the solution an 3nn3n for n0

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The characteristic equation as an alternative
representation

• In math, we often construct alternative
  representations of problems or structures, along
  with reversible one-to-one mappings between
  different representations.
• Then we can work with one such representation and
  transfer the results to others.
• One such mapping is between linear homogeneous
  recurrence relations and polynomials.

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Linear homogeneous recurrence relations of degree
k

• Characteristic equation
• rk c1rk-1 c2rk-2 ... ck 0
• Theorem 3 If the above characteristic equation
  has k distinct roots r1, r2, ..., rk, then an
  is a solution of the recurrence if and only if an
  b1r1n b2r2n ... bkrkn for all n0, where
  b1, b2, ..., bk are constants

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Example

• an 6an-1 11an-2 6an-3
• Initial conditions
• Characteristic equation
• Factoring
• Characteristic roots
• From the initial conditions we obtain
• b11, b2-1, b32
• Therefore,

a02, a15, a215
r3 6r2 11r 6 0
(r 1)(r 2)(r 3) 0
r1 1, r2 2, r3 3
an 11n 12n 23n 1 2n 23n
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Allowing multiple roots

• Theorem 4 If the above characteristic equation
  has t distinct roots r1, r2, ..., rt with
  multiplicities m1, m2, ..., mt (such that mi1
  and m1m2...mt k) then an is a solution of
  the recurrence if and only if
• for all n0, where bi,j are constants

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Reading

• Section 7.2 (Example 4 to Theorem 4)

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Review Problems

• Solve the recurrence relation an 4an-1 4an-2
  with initial conditions a0 3 and a1 8.
• Which recurrence relation has characteristic
  roots 2, 2, and -1?
• Find the general form of the solution of the
  recurrence relation an 8an-2 16an-4.