Big Oh Notation

Question

Answer 1

It is clear that here O(nlogn) is the biggest fn.

Confusion is between f(n) and g(n) who is bigger?

Take an asymptoticaly larger value

for i.e.  n=>$2^{10^{9}}$

now f(n)= $2^{100}$ and g(n) = $10^{9}$ =$2^{30}$

Hence g(n)<f(n)<h(n)

So finally Option D is Correct ans.

Comments

I checked it with n=10^99  but for that f(n) is coming 1 and g(n)=328 so how a larger value i can choose?

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Asymptotic means very very large Value..So We may take 2 to the power values due to log in g(n) so that we can cancel 2.

And here I took 2^10^9 bcz n^10^-7 was there ..

Answer 2

Take logarithm of all functions and then compare.

f = 0.0000001 * log(n)

g = log( log(n) )

h = log(n) + log( log(n) )

So, order is g< f <= h (equal to is used bacuse we are considering asymptotic notation)

So, option (D) is true.

Answer 3

In your question ,

$h\left ( n \right )> g\left ( n \right )> f\left ( n \right )$

$\Rightarrow 0\leq f\left ( n \right )\leq c*g\left ( n \right )\Rightarrow f\left ( n \right )=O\left ( g\left ( n \right ) \right )$

$\Rightarrow0\leq f\left ( n \right )\leq c*h\left ( n \right )\Rightarrow f\left ( n \right )=O\left ( h\left ( n \right ) \right )$

Answer C