Use the definition of Θ -notation to show that n 2 + 100 n + 88 is Θ ( n 2 ) .

To prove:

n2+100n+88 is Θ(n2).

Given information:

n2+100n+88.

Concept used:

Let f and g be real-valued functions defined on the same set of nonnegative integers, with g(n)≥0 for every integer n≥r, where r is a positive real number. Then f is of order g, written f ( n ) is Θ ( g ( n )) ( f of n is big-Theta of g of n ), if and only if, there exist positive real numbers A, B and k≥r such that

1. f is of order at least g, written f ( n ) is Ω ( g ( n )) ( f of n is big-Omega of g of n ), if and only if, there exist positive real numbers A and a≥r such that
   Ag(n)≤f(n) for every integer n≥a.
2. f is of order at most g, written f ( n ) is O ( g ( n )) ( f of n is big- O of g of n ), if and only if, there exist positive real numbers B and b≥r such that
   0≤f(n)≤Bg(n) for every integer n≥b.

Proof:

Firstly, to show that n2+100n+88 is Ω(n2), we need to show that n2+100n+88 is greater than or equal to a positive multiple of n2 for all values of n that are sufficiently large. Now

n2≤n2+100n+88 for every integer n≥1

Because when n≥1, 100n+88 is positive. Thus, we can let A=1 and a=1 to obtain

An2≤n2+100n+88 for every integer n≥a ,

Concluding by definition of Ω -notation, that n2+100n+88 is Ω(n2).

Secondly, to show that n2+100n+88 is O(n2), we need to show that n2+100n+88 is greater than or equal to 0 and less than or equal to some positive multiple of n2 for all values of n that are sufficiently large. First note that because all terms of n2+100n+88 are positive,

0≤n2+100n+88 for every integer n≥1