Use Theorems 11.2.5-11.2.9 and the results of exercises 15-17, 40, and 41 to justify the statements in 43-45.

45. ⌊ n 2 ⌋ + 4 n + 3 is Θ ( n )

To show:

That the polynomial ⌊n2⌋+4n+3 is θ(n).

Given:

The polynomial ⌊n2⌋+4n+3.

Formula used:

The order of a polynomial is given by,

If m is any integer with m≥0 and a1,a2,...,am are real numbers with am>0, then amnm+am−1nm−1+...+a1n+a0 is θ(nm)

Proof:

Consider p(n)=⌊n2⌋+4n+3.

We can define p(n) for all n≥0 ,

p(n)={⌊ n 2 ⌋+4n+3 when n is even⌊ n 2 ⌋+4n+3 when n is odd

We can obtain that n2 is an integer when n is even. Hence, ⌊n2⌋=n2

When n is odd, n−12 is the nearest even integer such that n−12≤n2