Homework Solution: Arrange the functions below in non-decreasing order such that if f_i appears before f_j then f_i(n) elementof O(f_j(n))….

    1. Arrange the functions below in non-decreasing order such that if f, appears before f then fi(n) O(J,(n)). [15 points] f1(n) = 1020 ½(n) = (lg n)4 fs(n) = 4n (n) = n lg n fs(n) = n. 100n2 fe(n) = n + lg n fr(n) = lg lg n fs(n) = no.1 Jo(n) = lgn5
    Arrange the functions below in non-decreasing order such that if f_i appears before 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

    Expert Answer

     
    Solution: We know that, f(n) epsilon O(g(n)), if f(n) <= c*g(n), where c is a positive constant.

    1. Arrange the employments beneath in non-decreasing prescribe such that if f, appears antecedently f then fi(n) O(J,(n)). [15 points] f1(n) = 1020 ½(n) = (lg n)4 fs(n) = 4n (n) = n lg n fs(n) = n. 100n2 fe(n) = n + lg n fr(n) = lg lg n fs(n) = no.1 Jo(n) = lgn5

    Arrange the employments beneath in non-decreasing prescribe such that if f_i appears antecedently 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

    Expert Counterpart

     

    Solution:

    We recognize that, f(n) epsilon O(g(n)), if f(n) <= c*g(n), where c is a direct continuous.

    f1(n) is lowest of them whole accordingly 10^20 is approximately equipollent to continuous (Irrespective of appraise 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, ce n=2^512, 512^4 gain be the outcome

    then comes f8(n)= n^0.1, we can obstruct the appraise the employment is unresisting by putting a very capacious appraise of n (2^512)

    (2^512)^0.1= 2.5866387e+15

    then f6(n)= n + lg n

    then f4(n)= n lg n

    then comes f5(n)= n^3-100*n^2,

    and at latest f3(n)= 4^n,

    So the prescribe gain go enjoy this,

    f1(n), f7(n), f9(n), f2(n), f8(n), f6(n), f4(n), f5(n), and f3(n).