Last time I had some questions on Big-Oh proofs, this set of problems is similar, I was wondering if the different notation still means the same thing, except now f(n) is always above 0? So "there exists some c so that if n is above the threshold B, then no matter what n is cg(n) will always be equal or greater than f(n)"?
I'm not really familiar with the new notation.
Thanks.
I'm not really familiar with the new notation.
Thanks.