Lecture Notes, 22 Nov 2010

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§§7.1–7.2: Recurrence Relations (cont'd)

1. Thm 1 (p. 462):
   Let C₀, C₁ ∈ N be constants.

   Let a₀ = C₀ and a₁ = C₁ be the initial conditions for a recurrence relation;
   i.e., the first 2 terms of a sequence.

   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 sequence defined recursively
     in terms of initial conditions and a recurrence relation
     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

   ---

2. Procedure (i.e., algorithm) for solving a (linear homogeneous) recurrence relation
   (of degree 2 with constant coefficients):
   • 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. Use:
                 –b + √(b² – 4ac)
         r₁ = ______________
                            2a
         and

                 –b – √(b² – 4ac)
         r₂ = ______________
                            2a

         where ar² + br + c = 0

         i.e.)

         a = 1
         b = –c₁
         c = –c₂

         ∴

                 –c₁ + √(c₁² – 4c₂)
         r₁ = ______________
                            2
         and

                 –c₁ – √(c₁² – 4c₂)
         r₂ = ______________
                            2

   3. Find α₁, α₂ such that a_n = α₁r₁ⁿ + α₂r₂ⁿ:
      • r₁ & r₂ were computed in step 2
      1. Use initial conditions to produce 2 simultaneous eqns in 2 unknowns:
         1. a₀ = α₁r₁⁰ + α₂r₂⁰ = α₁ + α₂ (!)

         2. a₁ = α₁r₁¹ + α₂r₂¹ = α₁r₁ + α₂r₂

      2. Solve these for α₁ & α₂
         • Note: α₁ = a₀ – α₂ (from 3(a)(i))

           ∴ a₁ = (a₀ – α₂)r₁ + α₂r₂ (from 3(a)(ii))
                     = a₀r₁ – α₂r₁ + α₂r₂
                     = a₀r₁ + α₂(r₂ – r₁)

           ∴ α₂(r₂ – r₁) = a₁ – a₀r₁

           ∴ α₂ = (a₁ – a₀r₁) / (r₂ – r₁)

           • …which is why r₁ ≠ r₂ !

           ∴ α₁ = a₀ – α₂ = a₀ – (a₁ – a₀r₁) / (r₂ – r₁)
                     = (a₀r₂ – a₀r₁ + a₀r₁ – a₁) / (r₂ – r₁)
                     = (a₀r₂ – a₁) / (r₂ – r₁)

   • O/P: explicit formula for a_n, namely:
     a_n = α₁r₁ⁿ + α₂r₂ⁿ

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3. As an example, let's solve the Fibonacci recurrence relation:
   1. f₀ = 0
      f₁ = 1

      f_n = f_n–1 + f_n–2

      0,1,1,2,3,5,…

   2. Solution:
      1. Char Eqn: r² – c₁r – c₂ = 0
         • What are c₁, c₂ s.t. f_n = c₁f_n–1 + c₂f_n–2?

         • Answer: c₁ = c₂ = 1

         • ∴ char eqn is: r² – r – 1 = 0
         Digression:
         • Consider a "golden rectangle":
           

           —allegedly the most aesthetically pleasing rectangle:

           • The front of the Parthenon has the shape of a golden rectangle.

           • Index cards come in golden rectangles: 3×5; 5×8 (note the Fibonacci numbers!)

         • The "golden ratio" (see the picture of the golden rectangle, above) is:
            r + 1      r
           _____ = _
               r         1

           ∴ r² = r + 1
           ∴ r² – r – 1 = 0  [does that look familiar?]

      2. Solve for r:
         • r = (1 ± √(1 + 4))/2 = (1 ± √5)/2

         • This is (these are?) the golden ratio(s):
           • √5 = 2.236067977…
           • (1+√5)/2 =   1.6180339885…
           • (1–√5)/2 = –0.6180339885…
           • Also: the reciprocal of (1+√5)/2 = 0.6180339885…
             • Weird!

      3. Find α₁, α₂ s.t. f_n = α₁r₁ⁿ + α₂r₂ⁿ :
         • Let r₁ = (1+√5)/2 & r₂ = (1–√5)/2

         • f₀ = 0 = α₁r₁⁰ + α₂r₂⁰ = α₁ + α₂

         • f₁ = 1 = α₁r₁¹ + α₂r₂¹

                     = α₁((1+√5)/2) + α₂((1–√5)/2)

         • Solve:
           • 0 = α₁ + α₂
           • ∴ α₁ = –α₂

           • 1 = α₁((1+√5)/2) + α₂((1–√5)/2)

           • ∴ 1 = –α₂((1+√5)/2) + α₂((1–√5)/2)

                     = α₂( (–1–√5)/2 + (1–√5)/2 )

                     = α₂( (–1–√5+1–√5)/2 )

                     = α₂( (–2√5)/2 )

                     = –α₂√5

           • ∴ α₂ = –1/√5

           • ∴ α₁ =   1/√5

      4. ∴ f_n = (1/√5)((1+√5)/2)ⁿ – (1/√5)((1–√5)/2)ⁿ
         • See lecture notes for 11/15/2010, §VI !

   ---

4. Another example:
   1. A long time ago, I discovered that if you fold a piece of paper in ½, then fold the top half back, then fold it ¼ of the way, then fold it back and fold it 3/8 of the way, and so on (see Fig. 1, below), the edge of the paper winds up 1/3 of the way from the edge!
      1. More precisely, fold it according to the following directions:
         

      2. 
         

      3. The question is: Where does B end up?

         And the answer is: 1/3 of the way from the edge!

         The puzzle is: How does a sequence of foldings-in-half yield the number 1/3?

   2. For the use of continuous math (i.e., limits) to prove this, see:
      • "Paper Folding and Convergent Sequences" (Rapaport 1974)

   3. Here's the sequence of folds, expressed recursively:
      a₀ = 0 (paper begins unfolded, at one edge, or 0")
      a₁ = ½ (first fold moves the edge ½, to middle of paper)

      a_n = (a_n–1 + a_n–2)/2

      (each subsequent fold moves the edge to the average of its previous two positions)

          = ½a_n–1 + ½a_n–2

      • 0, ½, ¼, 3/8, 5/16, 11/32, …

      • The curious feature is that each term in the sequence has a denominator that is a power of 2,
        yet the limit of the folds is 1/3.

        Where did the "3" come from?

   4. Let's solve this recurrence relation:
      • I/P: a_n = ½a_n–1 + ½a_n–2
      1. Char Eqn: r² – ½r – ½

      2. Solve for r:
         • r = (½ ± √(¼ + 4*½))/2
             = (½ ± √(¼ + 2))/2
             = (½ ± √(9/4))/2
             = (½ ± 3/2)/2
             ∈ {1, –½}

         • i.e.) r₁ = 1 & r₂ = –½

      3. Solve for α:
         • a₀ = 0 = α₁ + α₂

           ∴ α₂ = –α₁

         • a₁ = ½ = 1*α₁ + (–½*α₂) = α₁ – α₂/2

           ∴ ½ = α₁ + α₁/2 = 3α₁/2

           ∴ 1 = 3α₁

           ∴ α₁ = 1/3  [there's the 1/3!!!]

           ∴ α₂ = –1/3

      • O/P:
        a_n = α₁r₁ⁿ + α₂r₂ⁿ
             = (1/3)*1ⁿ + (–1/3)*(–½)ⁿ =      1/3 – (1/3)*(–½)ⁿ      = (1/3)(1 – (–½)ⁿ)
      • (As n gets bigger, the –½ⁿ gets smaller and smaller;
        in the limit it goes to 0, so the limit of the sequence is 1/3)

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5. 1. A final example, to be continued next time…

      Consider this sequence:

      a₀ = C₀
      a₁ = C₁

      a_n = 3a_n–1 – 2a_n–2, ∀ n ≥ 2

      Try it! Use the algorithm to show that a_n = 2ⁿ – 1, as suggested in the table in the previous lecture.

      (answer will be given in the next lecture)

Next lecture…