I'm having trouble with this one problem
9n <= cn^3
basically I can get down to
9/c <= n^2
But how do I solve the rest?
definition of little o
is
we say f(x)=o(g(x))
.
let f(x)=9*x and g(x)=c*x^3 where c is a positive constant. when x tends to infinity, f(x)/g(x) tends to 0.so we can say f(x)=o(g(x))
.
asyptotic notations are applicable for sufficiently large value of n.so for large value of n
9n << cn^3
for all c>0.