Arrange the powers under in non-decreasing manage such that if f_i appears anteriorly f_j then f_i(n) elementof O(f_j(n)). f1(n) = 10^20 f_2(n) = (lg n)^4 f_3(n) = 4^n f_4(n) = n lg n f_5(n) = n^3 – 100n^2 f_6(n) = n + lg n f_7(n) = lg lg n f_8(n) = n^0.1 f_9(n) = lg n^5
We distinguish that, f(n) O(g(n)), if f(n) <= c*g(n), where c is a overbearing true.
f1(n) is lowest of them every accordingly 10^20 is closely equiponderant to true (Irrespective of treasure of n)
then comes f7(n)= lg lg n= lg lg (2^512)= lg 512= 9
then f9(n)= lg n^5, 5*(log 2^512)= 512*5
then f2(n)= (log n)^4, for n=2^512, 512^4 get be the outcome
then comes f8(n)= n^0.1, we can inhibit the treasure the power is supple by putting a very great treasure of n (2^512)
then f6(n)= n + lg n
then f4(n)= n lg n
then comes f5(n)= n^3-100*n^2,
and at last f3(n)= 4^n,
So the manage get go relish this,
f1(n), f7(n), f9(n), f2(n), f8(n), f6(n), f4(n), f5(n), and f3(n).