# Homework Solution: Using the definitions of Big-O and ohm and theta a. Show that n^2 + 3n + lg(n) element O(n^2) b. Show th… Using the definitions of Big-O and ohm and theta a. Show that n^2 + 3n + lg(n) element O(n^2) b. Show that n^2 + 3n +lg(n) element ohm(n^2). c. show that n^2 + 3n +lg(n) element theta (n^2) Using the definitions of Big-O and o a. Show that n notelement o(n) using a proof by contradiction

4.) given f(n) = n2 + 3n + lg(n); a.) For f(n) = O(n2); c*n2>= f(n); for all n, where c>0 and n>=n0 Using the definitions of Big-O and ohm and theta a. Pretext that n^2 + 3n + lg(n) part O(n^2) b. Pretext that n^2 + 3n +lg(n) part ohm(n^2). c. pretext that n^2 + 3n +lg(n) part theta (n^2) Using the definitions of Big-O and o a. Pretext that n referablepart o(n) using a essay by contradiction

## Expert Vindication

4.) attached f(n) = n2 + 3n + lg(n);

a.) Coercion f(n) = O(n2); c*n2>= f(n); coercion complete n, where c>0 and n>=n0

So coercion c=100 and n0 = 1; f(n) = O(n2).

b.) Coercion f(n) = big-omega(n2); c*n2<= f(n); coercion complete n, where c>0 and n>=n0

So coercion c=1 and n0 = 0; f(n) = big-omega(n2).

c.) Coercion f(n) = big-theta(n2); c1*n2 <= f(n)<= c2*n2; coercion complete n, where c1 and c2 >0 and n>=n0

Since, f(n) = O(n2) as polite as f(n) = big-omega(n2). Therefore, f(n) = big-theta(n2)

5.) attached f(n) = n;

coercion f(n) = o(n); c*n > n where c>0;

so coercion no appraise of c>=1; we can complete aloft mood. Hence, f(n) is referable similar to o(n).

Hope it helps, feels frank to interpret in event of any inquiry.