Question #220967

(a) Suppose f and g are functions whose domains are subsets of z+, the set of positive integers. Give the definition of 'f is O\Omicron(g)'


(b) Use the definition of 'f is O\Omicron(g)' to show that:

(I) 2n+27 is O\Omicron(3n)

(ii) 5n is not O\Omicron(4n)


1
Expert's answer
2021-07-28T09:58:57-0400

Let ff and gg are functions whose domains are subsets of Z+\Z_+, the set of positive integers. We say that ff is O(g)O(g) if there exist a positive real number cc and a positive integer n0n_0 such that f(n)cg(n)f(n)\le cg(n) for all nn0.n\ge n_0.


(b) Let us use the definition of 'f is O\Omicron(g)' to show that the following statements.


(I) Since 2n+27<3n+273n=283n2^n+27<3^n+27\cdot 3^n=28\cdot 3^n for each positive integer n,n, we put c=28c=28 and conclude that 2n+272^n+27 is O(3n).O(3^n).


(ii) Let us show that 5n5^n is not O(4n).O(4^n). Let us prove using the method by contradiction. Suppose that there exist a positive real number cc and a positive integer n0n_0 such that 5nc4n5^n\le c\cdot 4^n for all nn0.n\ge n_0.

It follows that (54)nc(\frac{5}{4})^n\le c for all nn0.n\ge n_0. Since limn+(54)n=+,\lim\limits_{n\to+\infty}(\frac{5}{4})^n=+\infty, we conclude that there exist mZ+m\in\Z_+ such that (54)n>c(\frac{5}{4})^n>c for all nm.n\ge m. This contradiction proves the statement.


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