More Recurrence - Discrete Mathematical Structures - Lecture Slides, Slides of Discrete Mathematics

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Chapter 8

Recursion

More Recurrence

Example

• State whether each is a second-order linear

homogenous recurrence relation.

• a k = 3a k-1 + 2a k-

• b

k

= b

k-

+ b

k-

+ b

k-

• d

k

= d

2 k-

+ dk-

* dk-

• e

k

= 2e

k-

• f

k

= 2f

k-

Distinct Roots

• Lemma 8.3.

– Let A and B be real numbers. A recurrence relation

of the form a

k

= Aa

k-

+ Ba

k-

is satisfied by the

sequence 1, t, t

, t

, … , t

n

, … where t is non-zero

real number if, and only if, t satisfies the equation

t

– At – B = 0.

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. a (^) k = a (^) k-1 + 2a (^) k-
• Find all sequences that satisfy 1, t, t 2 , t 3 , … , t n , …
• Solution
  • t 2 – t - 2 = 0
  • t 2 - t - 2 = (t – 2)(t + 1), thus the values of t = 2, -
  • t can be: 1, 2, 2 2

3

, … , 2n OR 1, -

2

3

n

• This example demonstrates how to find two distinct

sequences that satisfy a given second-order homogenous

recurrence relation with constant coefficients.

Single Root Case

• Consider a

k

= Aa

k-

+ Ba

k-

for ints k≥2, but consider that the

characteristic equation t

– At – B = 0 has a single root.

• By Lemma 8.3.1 one sequence that satisfies the relation is:

1, r, r

, r

, … r

n

, … however another is: 0, r, 2r

, 3r

, … , nr

n

• To see this observe:

t

– At – B = (t – r)

= t

– 2rt – r

, A=2r B=-r

• s

n

= nr

n

for all ints n≥

• As

k-

+ Bs

k-

= A(k -1)r

k-

+ B(k -2)r

k-

= 2r(k -1)r

k-

+ -r

(k -2)r

k-

= 2(k -1)r

k

- (k -2)r

k

= (2k – 2 – k + 2) r

k

= kr

k

= s

k

Single Root

• Theorem 8.3.

– Suppose a sequence satisfies a recurrence relation

a

k

= Aa

k-

+ Ba

k-

for some real numbers A and B with B≠0 and for all

ints k≥2. If the characteristic eq t

– At – B = 0

has a single (real) root r then the sequence a

, a

, a

… satisfies the explicit formula

a

n

= Cr

n

+ Dnr

n

where C and D are the real numbers whose values are

determined by the values of a

and any other known

value of the sequence.

Example

• Suppose b 0 , b 1 , b 2 … satisfies the recurrence relation b k = 4b k- - 4b k- 2 for all ints k^ ≥ 2 with initial conditions b^0 = 1 and b 1 = 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 t 2 - 4t + 4 = 0 has a single root r = 2 ( (t-2)(t-2) )
  • bn = C 2 n + Dn n
  • to find C and D use initial conditions
    • b 0 = 1 = C 2^0 +D(0)2^0 => C = 1
    • b 1 = 3 = C 2 1 +D(1)2^1 => 2C + 2D = 3 (sub C = 1 from above)
    • 3 = 2(1) + 2D => D = ½
  • Hence, bn = 2 n + ½ n n for all ints n≥