Lecture Notes, 19 Nov 2010

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§§7.1–7.2: Recurrence Relations

1. Recurrence Relations:
   1. A sequence can be defined in two different ways:
      1. non-recursively ("explicitly"),
         in terms of its current I/P:
         a_n = f(n)

      2. recursively,
         by giving initial conditions (first few terms)
         a₀ = C₀, …, a_n–1 = C_n–1

         & a recurrence relation that defines the sequence in terms of its previous O/P:

         a_n = h(a₀ = C₀, …, a_n–1)

   2. Question: Given initial conditions & recurrence relation a_n = h(a₀ = C₀, …, a_n–1),
      (how) can we compute the explicit formula f(n)?
      • This is called "solving" the recurrence relation.

   3. E.g.)
      1. initial conditions:
         a₀ = C₀
         a₁ = C₁

         recurrence relation:

         a_n = 3a_n–1 – 2a_n–2, ∀ n ≥ 2

      2. Lots of different sequences share this pattern,
         differing only in their initial conditions
         1. Given the initial conditions,
            we can compute the nth term (n ≥ 2)
            without knowing what the function does to its I/P!

         2. We compute it on the basis of what it did:
            we compute it on the basis of what its previous O/P was!

         3. Here are some examples:
            (initial conditions are in the first 2 rows;
            last row shows "explicit" "solution",
            i.e.) def in terms of I/P)

            a0	0	0	1	1	1	2	2
            a1	0	1	0	1	2	1	2
            a2	0	3	–2	1	4	–1	2
            a3	0	7	–6	1	8	–5	2
            a4	0	15	–14	1	16	–13	2
            a5	0	…	–30	1	…	…	…
            …	…	…	…	…	…	…	…
            an	0	2n–1	2–2n	1	2n	3–2n	2

   4. What good are recurrence relations?
      1. They describe similar patterns of growth,
         based on differing initial conditions or "seeds"

      2. E.g.) compound interest:
         • 2 people deposit different amounts of $ in same bank;
           ∴ same recurrence relation computes their interest;

           But the actual interest depends on their initial deposit!

   ---

2. So the question is: How to "solve" a recurrence relation.
   1. Def:
      A linear, homogeneous, recurrence relation of degree 2
      with constant coefficients
      is_def
      a recurrence relation of the form
      a_n = c₁a_n–1 + c₂a_n–2

      where c₁, c₂ ∈ R & c₂ ≠ 0

      1. See text for complete def of linear homogeneous recurrence relation of degree k with constant coefficients.
      2. "linear": no exponents
      3. "homogeneous": all terms are multiples of the a_i
         • Note: pronounced "homoJEENee-us",
           not "hoMOJenus", with 5 syllables

      4. "constant coefficients":
         they are not functions of n;
         they are constants

   2. This determines a family of sequences
      that differ only in their initial conditions.
      • E.g.) The Fibonacci recurrence relation:
        f_n = f_n–1 + f_n–2

        is a linear homegeneous recurrence relation of degree 2
        and can have differing intial conditions,
        yielding different Fibonacci sequences:

        1. f₀ = 0 & f₁ = 1 yields:
           0,1,1,2,3,5,…

        2. f₀ = 1 & f₁ = 1 yields:
           1,1,2,3,5,…

        3. f₀ = 1 & f₁ = 2 yields:
           1,2,3,5,…

   3. They are solvable!

   ---

3. How do you solve them?
   1. The trick:
      • Given a₀=C₀, a₁=C₁, & a_n=c₁a_n–1+c₂a_n–2
        look for solutions of the form a_n = rⁿ, for constant r

   2. Why?
      1. Because, in the simplest case, when a_n=c₁a_n–1,
         the ratio of sucessive terms is constant;
         ∴ it's a geometric sequence

      2. Given the sequence
         a₀ = C₀
         a_n = c₁a_n–1

         we have:

         a₀ = C₀
         a₁ = c₁C₀
         a₂ = c₁²C₀
         …
         a_n = c₁ⁿC₀
         ∴ a_n = a₀c₁ⁿ

   3. a_n = rⁿ is a solution
      (i.e., an "explicit", non-recursive formula)
      for a_n = c₁a_n–1 + c₂a_n–2
      ↔
      (a_n =) rⁿ = c₁r^n–1 + c₂r^n–2 (by substituting rⁿ for a_n)
      ↔
      
      rⁿ         c₁r^n–1 + c₂r^n–2
      ___ = _____________ (for r ≠ 0)
      r^n–2           r^n–2
      
      
      ↔
      r² = c₁r + c₂
      ↔
      r² – c₁r – c₂ = 0 [the characteristic equation of the recurrence relation]
      ↔
      r is a solution of this equation [the characteristic root]
      (i.e., makes the equation come out T)

   ---

4. Thm 1 (p. 462):
   • In a theorem, you have to say where everything comes from;
     i.e., you must give their data types.

   Let C₀, C₁ ∈ N be constants.

   Let a₀ = C₀ and a₁ = C₁ be the initial conditions of a recurrence relation.

   Let c₁, c₂ ∈ R be such that a_n = c₁a_n–1 + c₂a_n–2 is the recurrence relation.

   Let r₁ ≠ r₂ be 2 distinct roots of the "characteristic equation"

   r² – c₁r – c₂

   of the recurrence relation. Then:

   
   (∃α₁, α₂ ∈ R)(∀n ∈ N)[a_n = α₁r₁ⁿ + α₂r₂ⁿ]
   • i.e.) the recurrence relation for the nth term
     can be computed non-recursively
     using the formula in terms of α_i and r_i

   • This is a non-constructive existence claim!
     • We need an algorithm to show how to find the α_i

   ---

5. (Outline of) procedure (i.e., algorithm)
   for solving a (linear homogeneous) recurrence relation
   (of degree 2 with constant coefficients):
   • Details will be given in the next lecture.
   • I/P: recurrence relation a_n = c₁a_n–1 + c₂a_n–2
   1. Set up the characteristic equation:
      r² – c₁r – c₂

   2. Solve the characteristic eqn for r₁, r₂
      1. if r₁ = r₂,
         then begin O/P "no solution"; halt end
         else goto (2b)

      2. Find α₁, α₂ such that a_n = α₁r₁ⁿ + α₂r₂ⁿ:
         1. Use initial conditions to produce 2 simultaneous eqns in 2 unknowns:

         2. Solve these for α₁ & α₂
      • O/P: explicit formula for a_n, namely:
        a_n = α₁r₁ⁿ + α₂r₂ⁿ

Next lecture…