Asymptotic-notations

Question

f(n)=2(log₂ n)² , g(n)=log₂n+1

How to give relation between them?.

Comments

For a value of n greater than or equal to 3, f(n) will be greater than g(n).

g(n)<=c*f(n), c=1 , n>=2

So, g(n)=O(f(n))

Is this the kind of relation you wanted?

---

@sakharam is right .... Lets expand this ->

Take any big value in term of power of 2 (Bcz ans get easily)

So, let suppose n=1024.

Put value in both eq.

$f(n)=2(logn)^2 = 2(log1024)^2$

                         = 2(log2^10)^2

                         = 2*10^2  = 200

$g(n)=(logn) + 1 = (log1024)+1$

                           10 + 1 = 11

So f(n) is greater then g(n).

We can say that g(n) =  O(f(n))

---

Thanks

Answer 1

g(n) = O(f(n))

Comments

How this relation is true? I think g(n) should be O(f(n))

---

sorry, it's g(n)=O(f(n))