Recursively Defined Sequences - Discrete Mathematical Structures - Lecture Slides, Slides of Discrete Mathematics

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Chapter 8

Recursion

Recursively Defined Sequences

Defining Recursion

• Recursion requires
  • recurrence relation relates later terms in the sequence to earlier terms and
  • initial conditions, values of the first few terms of the sequence.
• Example b0, b 1 , b 2 , … For all integers k>= 2,
  1. b (^) k = bk-1 + bk-2 (recurrence relation)
  2. b 0 = 1, b 1 = 3 (initial conditions) b 2 = b2-1 + b2-2 = b1 + b0 = 3 + 1 = 4 b 3 = b3-1 + b3-2 = b2 + b1 = 4 + 3 = 7 b 5 = b5-1 + b5-2 = b4 + b3 = 11 + 7 = 18

Recursion

• Definition:

• A recurrence relation for a sequence a 0 , a 1 , a 2 ,… is

a formula that relates each term a k to ceratin of its

predecessors a k-1, a k-2, … , a k-I, where I is an integer

and k is any integer greater than or equal to i.

• The initial conditions for such a recurrence

relation specify the values of a 0 , a 1 , a 2 ,…,a i-1, if I is

a fixed integer, or a 0 , a 1 , a 2 ,…,a m , where m is an

integer with m>=0, if i depends on k.

Equivalent Recursion

• There is more than one way to setup a recursive

sequence!

• for all k>=1, s (^) k = 3sk-1 – 1
• for all k>=0, s (^) k+1 = 3s (^) k – 1

• Are the two sequences equivalent?

• show the results of the sequence for terms:
  • starting at k = 1: s 1 = 3s 0 – 1, s 2 = 3s 1 – 1, s 3 = 3s 2 – 1
  • starting at k = 0: s (^) 0+1 = 3s 0 – 1, s (^) 1+1 = 3s 1 – 1, …
• change first term to second by adjusting first k
  • for all k>=1, s (^) k = 3s (^) k-1 – 1 => k start at 0 (k≥0) sk+1 =3s (^) k – 1
  • same as the second form.

Example

• Show that sequence given satisfies a recurrence

relation:

• 1, -1!, 2!, -3!, 4!, …, (-1) nn!, …, for n≥0 is equivalent to sk = (-k)sk-1 for k≥
• General term of the seq sn starting with s 0 = 1, then sn = (-
  1. nn! for n≥
     • substitute for k and k-1: sk = (-1) kk! , sk-1 = (-1) k-1(k-1)! (-k)s (^) k-1 = (-k)[(-1) k-1(k-1)!] (s (^) k defined above) = (k)(-1)(-1) k-1(k-1)! = (k)(-1) k(k-1)! = (-1) k(k)(k-1)! = (-1) k^ k! = sk

Towers of Hanoi

• Problem statement: eight

disks with holes in the

center that are stacked

from largest diameter to

smallest on the first of

three poles. Move the

stacked disk from one pole

to another.

• Rules: a larger disk cannot

be placed on top of a

smaller disk at any time.

Towers of Hanoi

• Recursive Solution
  1. Transfer the top k-1 disks from pole A to pole B. (Note: k>2 requires a number of moves.)
  2. Move the bottom disk from pole A to pole C.
  3. Transfer the top k-1 disks from pole B to pole C. (Again, if k>2, execution of this step will require more than one move.)

Towers of Hanoi

1. mk = 2mk-1 + 1 (recurrence relation)

2. m 1 = 1 (initial condition)

m 2 = 2m 1 + 1 = 3 m 3 = 2m 2 + 1 = 7 m 4 = 2m 3 + 1 = 15 … m 6 = 2m 5 + 1 = 63

Fibonacci

• Leonardo of Pisa was the greatest mathematician

of the 13 th^ century.

• Proposed the following problem:

• A single pair of rabbits (male/female) is born at the

beginning of a year. Assuming the following

conditions:

1. Rabbit pairs are not fertile during their first month of life but thereafter give birth to one new male/female pair at the end of every month.
2. No rabbits die

• How many rabbits will there be at the end of the

year?

Fibonacci

• for n ≥1, Fn = num of pairs alive at month n
• F 0 = the initial number of pairs
• F 0 = 1
• F (^) k = F (^) k-1 + F (^) k-
• Fibonacci Sequence

1. F (^) k = F (^) k-1 + F (^) k-2 (recurrence relation)
2. F 0 = 1, F 1 = 1 (initial condition) F 2 = F 1 + F 0 = 1+1 = 2 F 3 = F 2 + F 1 = 2+1 = 3 F 4 = F 3 + F 2 = 3+2 = 5 … F 12 = F 11 + F 10 = 144 + 89 = 233

Compound Interest

• Compute compound interest on principle

value using recursive sequence.

• $100,000 principle, earning 4% per year, how

much would you have in 21 years.

amt in account end of a year

=

amt in account at end of previous year

+

interest earned on account during year

=

amt in account at end of previous year

(0.04)* amt in account previous year A (^) n = amt in account at end of year n A 0 = initial amount (principle) A (^) k = A (^) k-1 + (0.04)Ak- A (^) k = (1.04)Ak-

Example

• Given $10,000, How much will the account be

work at the end of one year with a interest

rate of 3% compounded quarterly?

• Pk = Pk-1 (1+0.0075) = Pk-1(1.0075), integers k≥
• P 0 = 10,
• P 1 = 1.0075P 0 =1.007510,000 =10,075.
• P 2 = 10,150.
• P 3 = 10,226.
• P 4 = 10,303.