When c = 2 it is impossible for f(n) to be less than c*n^3 ,you have mentioned this for n not greater than 4 which is again wrong as for any value f(n) which is 3*n^3 will be greater than c*n^3 where c = 2.

Hi Akarsh!

The lesson has been updated. Thank you for your feedback!

Alina Fatima | Developer Advocate

10^n is same as 2^n, but when i choose true, quiz says the answer is wrong.

Proof :

log(10^n) <= log(c2^n)

n log 10 < = logc + nlog2

log 10, log c, log 2 are all constant factors, rest that remains is n <=n which is true.

Hi Rohan,

10^{n } is not in O(2^{n}).

Let f(n) = 10^{n} and g(n) = 2^{n}.

Now,

f(n) = (10)^{n} = (2^{3.322})^{n}

= (2^{n})^{3.322}

= g(n)^{3.322}

If f(n) = g(n)^{3.322} then,

f(n) can’t be in O(g(n)).

Hope this clarifies everything! Let me know if you have any other comments/questions/feedback.

Regards,

Alina Fatima | Developer Advocate

Hi Alina,

Thanks for the details/clear/ with proof reply, really appreciate. This answers/clears by understanding.