Question

Consider the following function, select all the function and apply which one is f= 0(g), f= omega g & f = theta g. Explain it each and every step.

f = n! g = 2^n

f=(log n)^3 g = n

f= 5^n/2 g =2^n

f=logn! g= nlog n

f=3^(n+1) g= 5^n

f=n! g= 2^n

f=2^n g= 2^n/2

f=2n+logn g = n+(logn)^2

f=nsqrt(n) g= 5^(log_2 (n))

Answer 1

1
f = n! g = 2^n
for n > 2
   f(n) > g(n)

so f = Omega(g)

2.

f=(log n)^3 g = n

for n >=1
    f(n) < g(n)

So f = O(g)

3.

f= 5^n/2 g =2^n

for n >=1
    f(n) > g(n)

So f = Omega(g)

4.

f=logn! g= nlog n

n! is similar to nⁿ, so f = log nⁿ, = nlogn

So f = theta(g)

5

f=3^(n+1) g= 5^n

similar to 3

for n >= 3
g(n) > f(n)

so f(n) = O(g)

6

f=n! g= 2ⁿ

as said earlier n! ssimilar to nⁿ. We are quite aware of that nⁿ > 2ⁿ

so f = theta(g)

7

f=2^n g= 2^n/2

anytime
f(n) > g(n)

so f = theta(g)

8

f=2n+logn g = n+(logn)^2

f = 2n+logn =     n + n + log n

g = n + (logn)² = n + log n * log n

clearly n + logn > log n * log n

9
f=nsqrt(n) g= 5^(log_2 (n))

exponential series is always > quadratic series

so

f(n) < g(n)

f = O(g)