Chapter 8
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Chapter 8. Recursion. 8.3. More Recurrence. Second-Order Recurrence. Definition

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Chapter 8

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Chapter 8

Chapter 8

Recursion


Chapter 8

8.3

More Recurrence


Second order recurrence

Second-Order Recurrence

  • Definition

    • A second-order linear homogeneous recurrence relation with constant coefficients is a recurrence relation of the form,ak = Aak-1 + Bak-2 for all intsk ≥ some fixed int, where A and B are fixed real numbers and B≠0.

    • This is called a second order because expression is based on the previous two terms (ak-1 , ak-2).

    • Its linear ak-1 , ak-2 are separate terms of the first order.

    • Homogenous because are terms are of the same order.

    • Constant coefficient b/c A and B are fixed real numbers that do not depend on k.


Example

Example

  • State whether each is a second-order linear homogenous recurrence relation.

    • ak = 3ak-1 + 2ak-2

    • bk = bk-1 + bk-2 + bk-3

    • dk = d2k-1 + dk-1 * dk-2

    • ek = 2ek-2

    • fk = 2f k-1 + 1


Distinct roots

Distinct Roots

  • Lemma 8.3.1

    • Let A and B be real numbers. A recurrence relation of the form ak = Aak-1 + Bak-2 is satisfied by the sequence 1, t, t2, t3, … , tn, … where t is non-zero real number if, and only if, t satisfies the equation t2 – At – B = 0.


Characteristic equation

Characteristic Equation

  • Definition

    • Given a second-order linear homogenous recurrence relation with constant coefficients: ak = Aak-1 + Bak-2 for all ints k≥2,the characteristic equation of the relation is t2 – At – B = 0


Example1

Example

  • Use characteristic eq to find solutions to a recurrence problem.

    • Consider recurrence relation where kth term of a sequence equals the sum of the (k-1)st term plus twice the (k-2) term.ak = ak-1 + 2ak-2

    • Find all sequences that satisfy 1, t, t2, t3, … , tn, …

    • Solution

      • t2 – t - 2 = 0

      • t2 – t - 2 = (t – 2)(t + 1), thus the values of t = 2, -1

      • t can be: 1, 2, 22, 23, … , 2n OR 1, -12, -13, … -1n

  • This example demonstrates how to find two distinct sequences that satisfy a given second-order homogenous recurrence relation with constant coefficients.


Single root case

Single Root Case

  • Consider ak = Aak-1 + Bak-2 for ints k≥2, but consider that the characteristic equation t2 – At – B = 0 has a single root.

  • By Lemma 8.3.1 one sequence that satisfies the relation is: 1, r, r2, r3, … rn, … however another is: 0, r, 2r2, 3r3, … , nrn

  • To see this observe:t2 – At – B = (t – r)2 = t2 – 2rt – r2, A=2r B=-r2

  • sn= nrn for all ints n≥0

  • Ask-1 + Bsk-2 = A(k -1)rk-1 + B(k -2)rk-2 = 2r(k -1)rk-1 + -r2 (k -2)rk-2= 2(k -1)rk -(k -2)rk = (2k – 2 – k + 2) rk= krk = sk


Single root

Single Root

  • Lemma 8.3.4

    • Let A and B be real numbers and suppose the characteristic eq t2 – At – B = 0 has a single root r. Then the sequence 1, r1, r2, … , rn, … and 0, r, 2r2, … nrn, … both satisfy the recurrence relationak = Aak-1 + Bak-2 for all ints k≥2


Single root1

Single Root

  • Theorem 8.3.5

    • Suppose a sequence satisfies a recurrence relation ak = Aak-1 + Bak-2for some real numbers A and B with B≠0 and for all ints k≥2. If the characteristic eq t2 – At – B = 0 has a single (real) root r then the sequence a0, a1, a2, … satisfies the explicit formula an = Crn + Dnrnwhere C and D are the real numbers whose values are determined by the values of a0 and any other known value of the sequence.


Example2

Example

  • Suppose b0, b1, b2 … satisfies the recurrence relation bk = 4bk-1 – 4bk-2 for all intsk ≥ 2 with initial conditions b0 = 1 and b1 = 3.Find an explicit formula for the sequence.

  • Solution

    • sequences is of second-order linear homogenous recurrence relation with constant coefficients (A=4 and B=-4). The single-root condition is also met because the characteristic equation t2 – 4t + 4 = 0 has a single root r = 2 ( (t-2)(t-2) )

    • bn = C 2n + Dn2n

    • to find C and D use initial conditions

      • b0 = 1 = C 20 +D(0)20 => C = 1

      • b1 = 3 = C 21 +D(1)21 =>2C + 2D = 3 (sub C = 1 from above)

      • 3 = 2(1) + 2D => D = ½

    • Hence, bn = 2n + ½ n2n for all intsn≥2


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