Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 13

1. Show, by direct inference, that the function f(x) = 10 is Theta(g(x)) where 
g(x) = 1.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |10|
       = 10
       = 10*1
       = C*|1|		for C=10
       =< C*|g(x)|	for K is any value and C=10

Proving f(x) is Omega(g(x)):
|f(x)| = |10|
       = 10
       = 10*1
       = C*|1|		for C=10
       >= C*|g(x)|	for K is any value and C=10

2. Show, by direct inference, that f(x) = 3*x-7 is Theta(g(x)) where g(x)=x.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |3*x-7|
       = 3*x-7		for x>3
       =< 3*x
       = 3*|x|
       = C*|x|		for C=3
       = C*|g(x)|	for K=3 and C=3

Proving f(x) is Omega(g(x)):
|f(x)| = |3*x-7|
       = 3*x-7		for x>3
       >= 2*x		for x>7
       = 2*|x|
       = C*|x|		for C=2
       = C*|g(x)|	for K=7 and C=2

3. Show, by direct inference, that f(x) = x^2-x-1 is Theta(g(x)) where g(x)=x^2.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |x^2-x-1|
       = x^2-x-1	for x>2
       =< x^2
       = 1*|x^2|
       = C*|x^2|	for C=1
       = C*|g(x)|	for K=2 and C=1

Proving f(x) is Omega(g(x)):
|f(x)| = |x^2-x-1|
       = x^2-x-1	for x>2
       = (x^2)/2 + (x^2)/2 - x - 1 
       >= (x^2)/2       for x>4
       = (1/2)*|x^2|
       = C*|x^2|	for C=(1/2)
       = C*|g(x)|	for K=4 and C=(1/2)

4. Show, by direct inference, that f(x) = 5*log(x) is Theta(g(x)) where 
g(x)=log(x).

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |5*log(x)|
       = 5*log(x)	for x>2
       = C*log(x)	for C=5
       = C*|log(x)|
       =< C*|g(x)|	for K=2 and C=5

Proving f(x) is Omega(g(x)):
|f(x)| = |5*log(x)|
       = 5*log(x)	for x>2
       >= (4)*log(x)
       = C*log(x)	for C=4
       = C*|log(x)|
       = C*|g(x)|	for K=2 and C=4

5. Show, by direct inference, that f(x) = ceiling(x) is Theta(g(x)) where 
g(x)=x.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |ceiling(x)|
       = ceiling(x)	for x>1
       =< 2*x
       = C*|x|		for C=2
       = C*|g(x)|	for K=1 and C=2

Proving f(x) is Omega(g(x)):
|f(x)| = |ceiling(x)|
       = ceiling(x)	for x>1
       >= x
       = C*|x|		for C=1
       = C*|g(x)|	for K=1 and C=1

6. Show, by direct inference, that f(x) = (x^2+1)/(x+1) is Theta(g(x)) where 
g(x)=x.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |(x^2+1)/(x+1)|
       = (x^2+1)/(x+1)	for x>1
       =< (x^2+1)/x
       =< (x^2+x^2)/x
       = 2*x
       = C*|x|		for C=2
       = C*|g(x)|	for K=1 and C=2

Proving f(x) is Omega(g(x)):
|f(x)| = |(x^2+1)/(x+1)|
       = (x^2+1)/(x+1)	for x>1
       >= (x^2+1)/(2*x)
       >= (x^2)/(2*x)
       = (1/2)*x 
       = C*|x|		for C=(1/2)
       = C*|g(x)|	for K=1 and C=(1/2)

7. Show, by direct inference, that 
   (f1(x) is O(g1(x))) and (f2(x) is O(g2(x))) 
   => (f1(x)*f2(x) is O(g1(x)*g2(x)))

Sol:
 (f1(x) is O(g1(x))) and (f2(x) is O(g2(x)))
=> {Definition of "O"}
 (Exist K1, C1, K2, C2,
  (All x>K1, |f1(x)| =< C1*|g1(x)|) and
  (All x>K2, |f2(x)| =< C2*|g2(x)|))
=> {Let K be max(K1,K2)}
 (Exist K, C1, C2,
  (All x>K, |f1(x)| =< C1*|g1(x)|) and
  (All x>K, |f2(x)| =< C2*|g2(x)|))
=> {Arithmetics}
 (Exist K, C1, C2,
  (All x>K, |f1(x)| =< C1*|g1(x)| and |f2(x)| =< C2*|g2(x)|)
=> {Arithmetics}
 (Exist K, C1, C2,
  (All x>K, |f1(x)|*|f2(x)| =< C1*C2*|g1(x)|*|g2(x)|)
=> {Arithmetics}
 (Exist K, C1, C2,
  (All x>K, |f1(x)*f2(x)| =< C1*C2*|g1(x)*g2(x)|)
=> {Let C be C1*C2}
 (Exist K, C,
  (All x>K, |f1(x)*f2(x)| =< C*|g1(x)*g2(x)|)
=> {Definition of "O"}
 (f1(x)*f2(x)) is O(g1(x)*g2(x))

8. Show, by direct inference, that f(x) = 4x^2-5x+3 is O(g(x)) where 
g(x) = x^2. 

Sol:
|f(x)| = |4x^2-5x+3|
       = 4x^2-5x+3		for x>1
       =< 4x^2+5x^2+3x^2
       = 12x^2
       = C*|x^2|		for C=12
       = C*|g(x)|

9. Show, by direct inference, that f(x) = (x+5)*log(3x^2+7) is O(g(x)) where 
g(x) = x*log(x).

Sol:
|f(x)| = |(x+5)*log(3x^2+7)|
       = (x+5)*log(3x^2+7)	for x>1
       =< (x+5x)*log(3x^2+7x^2)
       = 6x*log(10x^2)
       = 6x*log(x^3)		for x>10
       = 18x*log(x)
       = C*|x*log(x)|		for C=18
       = C*|g(x)|		for K=10 and C=18  

10. Show, by direct inference, that f(x) = (x^2+5*log(x))/(2x+1) is O(g(x)) 
where g(x) = x.

Sol:
|f(x)| = |(x^2+5*log(x))/(2x+1)|
       = (x^2+5*log(x))/(2x+1)	for x>1
       =< (x^2+5*x^2)/(2x+1)
       = (6*x^2)/(2x+1)
       =< (6*x^2)/(2x)
       = 3*x
       = C*|x|			for C=3
       = C*|g(x)|		for K=1 and C=3

11. Show, by direct inference, that f(n) = log(1) + log(2) + ... + log(n), 
where n is a positive integer, is O(g(n)) where g(n) = n*log(n).

Sol:
|f(n)| = |log(1) + log(2) + ... + log(n)|
       = log(1) + log(2) + ... + log(n)		for n>1
       =< log(n) + log(n) + ... + log(n)
       = n*log(n)
       = C*|n*log(n)|				for C=1
       = C*|g(n)|