Explanation Given: The polynomial n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 . Formula used: The order of a polynomial is given by, If m is any integer with m ≥ 0 and a 1 , a 2 , ... , a m are real numbers with a m > 0 , then a m n m + a m − 1 n m − 1 + ... + a 1 n + a 0 is θ ( n m ) Proof: Let p ( n ) be n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 . For all n ≥ 0 we can define p ( n ) , p ( n ) = { n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 when n is even n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 when n is odd For n is even, we can obtain that n 2 is an integer. Hence, ⌊ n 2 ⌋ = n 2 Also, when n is odd, n 2 is not an integer. But, n − 1 2 is an integer such that n 2 ≥ n − 1 2

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ISBN: 9780357097717

Author: EPP

Publisher: CENGAGE L

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Chapter 11.2, Problem 44ES

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That the polynomial n(n−1)2+⌊n2⌋+1 is θ(n2).

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Chapter 11.2, Problem 43ES

Chapter 11.2, Problem 45ES

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That the polynomial n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 is θ ( n 2 ) .

Solution Summary: The author explains that the polynomial n(n-1)2+lfloor.

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That the polynomial n(n−1)2+⌊n2⌋+1 is θ(n2).

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Chapter 11 Solutions

That the polynomial n ( n − 1 ) 2 + ⌊ n 2 ⌋ + 1 is θ ( n 2 ) .

Solution Summary: The author explains that the polynomial n(n-1)2+lfloor.

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To show: That the polynomial n(n−1)2+⌊n2⌋+1 is θ(n2).

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Chapter 11 Solutions

To show:

That the polynomial n(n−1)2+⌊n2⌋+1 is θ(n2).