Mohamed G. Gouda						     CS 313K
Fall 2012							     Homework 5

1. Consider the recurrence equation:
   T(1) = 2
   T(2) = 9
   T(n+2) = 2*T(n+1) + 3*T(n)		for n >= 1
Show, by induction, that
   T(n) =< 3^n				for n >= 1

Sol:
Let P(n) be the predicate T(n) =< 3^n. 

Base Case: n=1 and n=2:
P(1) <=> T(1) =< 3 <=> 2 =< 3 <=> T
P(2) <=> T(2) =< 9 <=> 9 =< 9 <=> T

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

 T
=> {Definition of T(n+2)}
 T(n+2) = 2*T(n+1) + 3*T(n)
=> {Induction Hypothesis P(n) and P(n+1)}
 T(n+2) =< 2*(3^(n+1)) + 3*(3^n)
 	= (2+1)*(3^(n+1))
 	= 3^(n+2)
=> {Definition of P(n+2)}
 P(n+2)

2. There are two ways to climb the steps of a stair: One step at a time and two 
steps at a time. Let T(n) denote the number of ways that can be used to climb 
n steps. Derive the recurrence equation for T(n).

Sol:
Clearly, T(1) = 1, T(2) = 2, T(3) = 3, T(4) = 5, ... The different ways to  
climb n steps are as follows: 

(1) First climb n-1 steps, using T(n-1) different ways, then climb the last 
step in one go. 

(2) First climb n-2 steps, using T(n-2) different ways, then climb the last two 
steps in one go.

Therefore, T(n) = T(n-1) + T(n-2).

Thus, the recurrence equation is:
      T(1) = 1
      T(2) = 2
      T(n+2) = T(n+1) + T(n)			for n >= 1 

3. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 5						(1)
      T(1) = 2						(2)
      T(n+2) = -10*T(n+1) - 25*T(n)	for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = -10*t - 25
t^2 + 10*t + 25 = 0
(t+5)^2 = 0
t1 = -5 and t2 = -5

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

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

From (2), (4), and (5):
2 = 5*(-5) + y*(-5)
y = -(27/5)						(6)

The closed equation from (4), (5), and (6):
T(n) = 5*(-5)^n  - (27/5)*n*(-5)^n

4. An amount of $10,000 is deposited in a savings account whose interest is 
5% per year. How much money will be in the account after 30 years? Derive 
your answer.

Sol:
Let T(n) denote the amount of money in the account after 30 years.
    T(0) = 10,000
    T(n+1) = (1.05)*T(n)

Thus,
    T(1) = (1.05)*T(0)
    T(2) = (1.05)^2 * T(0)
    ...
    T(30) = (1.05)^(30) * T(0)
    	  = (1.05)^(30) * (10,000)
	  = 43,219.42

Therefore, the account will have $43,219.42 after 30 years.

5. Show, by direct inference, that 
   (f(x) is O(g(x))) and (g(x) is O(h(x))) 
   => (f(x) is O(h(x)))

Sol:
 (f(x) is O(g(x))) and (g(x) is O(h(x))) 
=> {Definition of "O"}
 (All x > K1, |f(x)| =< C1*|g(x)|) and 
 (All x > K2,|g(x)| =< C2*|h(x)|) 
=> {Let K be max(K1,K2)}
 (All x > K, |f(x)| =< C1*|g(x)| and |g(x)| =< C2*|h(x)|)
=> {Arithmetics}
 (All x > K, |f(x)| =< C1*|g(x)| =< C1*C2*|h(x)|)
=> {Let C be C1*C2}
 (All x > K, |f(x)| =< C*|h(x)|)
=> {Definition of "O"}
 (f(x) is O(h(x)))

6. Show, by contradiction, that x^3 is not O(7 * x^2).

Sol:
 x^3 is O(7 * x^2)
=> {Definition of "O"}
 (Exist C and K, (All x > K, |x^3| =< C*|7 * x^2|))
=> {Arithmetics}
 (Exist C and K, (All x > K, |x| =< C*|7|))
=> {Choose any value x' that is larger than max(K,C*7)}
 (Exist C and K, (All x > K, |x| =< C*|7|) and 
 	      	 (Exist x' > K, |x'| > C*7))
=> F