Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 12

1. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 4						(1)
      T(1) = 3						(2)
      T(n+2) = 3*T(n+1) + 4*T(n)	for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = 3*t + 4
t^2 - 3*t - 4 = 0
(t+1)(t-4) = 0
t1 = -1 and t2 = 4

Expression for T(n):
T(n) = x*(-1)^n + y*(4)^n				(4)

From (1) and (4):
4 = x + y						(5)

From (2) and (4):
3 = -x + 4*y						(6)

From (5) and (6),
x = 13/5 and y = 7/5					(7)

The closed equation from (4) and (7):
T(n) = (13/5)*(-1)^n  + (7/5)*(4)^n

2. Use two methods, i.e. Poorman's Induction and Characteristic Polynomial, to 
derive the closed equation for the following recurrence equation:
       T(0) = 7	  	       (1)
       T(n+1) = 2*T(n)	       (2)

Sol:

Poorman's Induction Method
From (2),
T(n) = 2*T(n-1)
     = (2^2)*T(n-2)
     = (2^3)*T(n-3)
     ...
     = (2^i)*T(n-i)
     = {Choosing i to be n}
       (2^n)*T(0)
     = {From (1)}
       (2^n)*7
       
Characteristic Polynomial
The Characteris Polynomial from (2) is:
t = 2
This polynomial has one root t1 whose value is 2.

Expression for T(n):
T(n) = x*(t1)^n
     = x*(2)^n

From (1), we get
   7 = x*1

The closed equation is:
T(n) = 7*(2^n)

3. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 6						(1)
      T(1) = 8						(2)
      T(n+2) = 4*T(n+1) - 4*T(n)	for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = 4*t - 4
t^2 - 4*t + 4 = 0
(t-2)^2 = 0
t1 = 2 and t2 = 2

Expression for T(n):
T(n) = x*(2)^n + y*n*(2)^n				(4)

From (1) and (4):
6 = x	 						(5)

From (2), (4) and (5):
8 = 2*6 + 2*y
y = -2							(6)	

From (5) and (6),
x = 6 and y = -2					(7)

The closed equation from (4) and (7):
T(n) = 6*(2)^n  -  2*n*(2)^n

4. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 1						(1)
      T(1) = 1						(2)
      T(2) = 2						(3)
      T(n+3) = (-2)*T(n+2) + T(n+1) + 2*T(n) for n >= 0 (4)

Sol:
The Characteristic Polynomial from (4) is:
t^3 = -2*t^2 + t + 2
t^3 + 2*t^2 - t - 2 = 0
(t+1)(t-1)(t+2) = 0
t1 = -1, t2 = 1 and t3 = -2

Expression for T(n):
T(n) = x*(-1)^n + y*(1)^n + z*(-2)^n			(5)

From (1), (2), (3) and (5):
x = -1/2, y = 7/6, and z = 1/3				(6)

The closed equation from (5) and (6):
T(n) = (-1/2)*(-1)^n  + (7/6) + (1/3)*(-2)^n

5. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation of Lucas Numbers:
      T(0) = 2						(1)
      T(1) = 1						(2)
      T(n+2) = T(n+1) + T(n)		for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = t + 1
t^2 - t - 1 = 0

{Hint: The two roots are 
(-b + sqrt(b^2 - 4ac))/2a and 
(-b - sqrt(b^2 - 4ac))/2a.}

t1 = (1 + sqrt(5))/2 and t2 = (1 - sqrt(5)) / 2

Expression for T(n):
T(n) = x*((1 + sqrt(5))/2)^n + y*((1 - sqrt(5))/2)^n	(4)

From (1), (2), and (4):
x = 1 and y = 1						(5)

The closed equation from (4) and (5):

T(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n

6. Consider the recurrence equation of Lucas Numbers:
   T(0) = 2
   T(1) = 1
   T(n+2) = T(n+1) + T(n)	for n >= 0
Prove, by induction, that
   T(n) = F(n+1) + F(n-1)	for n >= 2
where F(n) is the n-th Fibonacci Number.

Sol:
Let P(n) be the predicate T(n) = F(n+1) + F(n-1).

Base Case: n=2 and n=3:
P(2) <=> T(2) = T(1)+T(0) = 1+2 = F(3)+F(1) <=> T
P(3) <=> T(3) = T(2)+T(1) = 3+1 = F(4)+F(2) <=> T

Induction Step: 
For n >= 2, P(n-1) and P(n) => P(n+1)

 T
=> {Definition of T(n+1)}
 T(n+1) = T(n) + T(n-1)
=> {Induction Hypothesis: P(n-1) and P(n)}
 T(n+1) = (F(n+1)+F(n-1)) + (F(n)+F(n-2))
=> {Definition of Fibonacci Numbers}
 T(n+1) = F(n+2) + F(n) 
=> {Definition of P(n+1)}
 P(n+1)

7. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 2						(1)
      T(1) = -4						(2)
      T(n+2) = 2*T(n+1) - T(n)		for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = 2*t - 1
t^2 - 2*t + 1 = 0
(t-1)^2 = 0
t1 = 1 and t2 = 1

Expression for T(n):
T(n) = x*(1)^n + y*n*(1)^n
     = x + y*n 	 					(4)

From (1) and (4):
2 = x	 						(5)

From (2), (4), and (5):
-4 = 2 + y
y = -6							(6)	

The closed equation from (4), (5), and (6):
T(n) = 2 - 6*n