1. (a) Find the best possible relationship using one of the notations: O, Ω, Θ, o, w, for the following pairs of functions: n3 + 7n1.5-3 and nlog8 – 10n°.5 + 7; nlgn and n11; 5″ and (5.1)”; 4″ and n!. Justify each answer (b) Function f(n) = n3 + 5000n-60000 when n 〈 50 and f(n) = nlog n+ 100n for n 〉 50. Write f(n) in asymptotic notation in the simplest possible form.

Solution:

a)

n^3+7(n^1..5)-3 and n^(log 8 -10(n^0.5)+7)

we already know how Big- O, Big-Omega, Theta works but here is an overview to go through once again,

By looking at functions we can observe that

n^3+7(n^1..5)-3 <= c1* ( n^(log 8 -10(n^0.5)+7))

and Please note that log2 8= log 2^3= 3 log 2= 3*1= 3, so n^log 8= n^3

n^3+7(n^1..5)-3 >= c2* (n^(log 8 -10(n^0.5)+7))

So we can say that

n^3+7(n^1..5)-3= $\Theta$(n^(log 8 -10(n^0.5)+7) )

Now,

n log n and n^1.1

Now we will compare by putting very large value of n and see the relation between these two

if n=2^256, then n log n= 2.9642775e+79 and

n^1.1= 5.8890708e+84

this means that n^1.1 is greater than n log n

Hence n log n= O(n^1.1).

5^n and 5.1^n

Only by looking at the function we can say that

5^n= o(5.1^n)

4^n and n!

We know that n! is almost equivalent to n^n

So here we can say that 4^n= o(n!)

b)

When we write generalized form of asymptotic notation then we only consider very large value of n, so the function which is taken at n>50 is going to be considered here,

f(n)= n log n + 100n

n log n + 100n <= c * (n log n )

for c= 2.

So, f(n)= O(n log n)

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