differentiate:f'= (cn-logn)'=c-1/n. With n>n0=1/c ->f'>0->increasing
function. Therefore, n>n0 ,lognn0'.
because logn<>omega(n)->logn<>theta(n)
On Tue, Feb 2, 2010 at 12:51 PM, Banoo wrote:
> hi all,
> can you help me solve the following question.
>
> Is log n = O(n)? Is log n = omega(n)? Is log
initial assignmnet: g()={n1,..,nd)
inter(i)
{
if(i<0) return;
for(int j=0;j wrote:
> Can you please help me with this one.
>
> write two algorithms that iterate over every index from (0,0,..,0)
> to (n1,n2,..,nd).Make one algorithm recursive and the other
> iterative.
>
> Th
