(a) Explanation Proof: A definition states that for two functions f and g , Then, f = o ( g ) can be written if there is an integer n 0 and a positive real number c , Such that, | f ( n ) | ≤ c | g ( n ) | for all n ≥ n 0 The triangle inequality states that the inequality holds | a + b | ≤ | a | + | b | for any real numbers a and b . One proposition states that the polynomial is the same as its higher power. In other words, Given that, A polynomial of degree t , which is f ( n ) = a t n t + a t − 1 n t − 1 + ... + a 0 , then, n ≥ n 0 , c = 2 a 1 n t = o ( f ( n ) ) . This proposition is proved by using the triangle inequality to show for any natural number n : f ( n ) ≤ | a t n t | + | a t − 1 n t − 1 | + ... + | a 1 n | + | a 0 | = | a t | n t + | a t − 1 | n t − 1 + ... + | a 1 | n + | a 0 | , As, n > 0 ≤ | a t | n t + | a t − 1 | n t + ... + | a 1 | n t + | a 0 | n t = ( | a t | + | a t − 1 | + ... + | a 1 | + | a 0 | ) n t ....... ( 1 ) From the above inequality it can be seen that | f ( n ) | ≤ c n t for any n ≥ n 0 = 1 , Where, c = | a t | + | a t − 1 | + ... + | a 1 | + | a 0 | . Therefore, the definition given at the beginning of this document, it can be seen that, f ( n ) = o ( n t ) . To prove that, n t = o ( f ( n ) ) : as f © − f , it can be assumed that, a t > 0 , So that, f ( n ) = a t n t + a t − 1 n t − 1 + a t − 2 n t − 2 + ... + a 0 = n t ( a t + a t − 1 n + | a t − 2 | n 2 + ... + a 0 n t ) ≥ n t ( a t − | a t − 1 | n − | a t − 2 | n 2 − ... − | a 0 | n t ) ...... ( 2 ) For large values of n, the following terms will get close to 0: f ( n ) = n t | a t − 1 | n | a t − 2 | n 2 < a t 2 t , ... , | a 0 | n t < a t 2 t − | a t − 1 | n > − a t 2 t , − | a t − 2 | n 2 > − a t 2 t , ... , − | a 0 | n t > − a t 2 t ....... ( 3 ) So that, The large n these terms will be smaller than the term a t 2 t , that is: | a t − 1 | n < a t 2 t , | a t − 2 | n 2 < a t 2 t , (b) To determine To prove: n 4 − 2 n 3 + 3 n 2 − 5 n + 7 © n 4

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

3rd Edition

ISBN: 9780134689555

Author: Edgar Goodaire, Michael Parmenter

Publisher: PEARSON

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Chapter 8, Problem 8RE

(a)

To determine

To prove: 3n3−5n2+2n+1©n3

(b)

To determine

To prove: n4−2n3+3n2−5n+7©n4

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Chapter 8, Problem 7RE

Chapter 8, Problem 9RE

Chapter 8 Solutions

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 8.1 - Prob. 1TFQ Ch. 8.1 - Prob. 2TFQ Ch. 8.1 - Prob. 3TFQ Ch. 8.1 - Prob. 4TFQ Ch. 8.1 - Prob. 5TFQ Ch. 8.1 - Prob. 6TFQ Ch. 8.1 - Prob. 7TFQ Ch. 8.1 - Prob. 8TFQ Ch. 8.1 - Prob. 9TFQ Ch. 8.1 - Prob. 10TFQ

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - Prob. 13E Ch. 8.1 - Prob. 14E Ch. 8.1 - Prob. 15E Ch. 8.1 - Prob. 16E Ch. 8.1 - Prob. 17E Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.2 - Prob. 1TFQ Ch. 8.2 - Prob. 2TFQ Ch. 8.2 - Prob. 3TFQ Ch. 8.2 - Prob. 4TFQ Ch. 8.2 - Prob. 5TFQ Ch. 8.2 - Prob. 6TFQ Ch. 8.2 - Prob. 7TFQ Ch. 8.2 - Prob. 8TFQ Ch. 8.2 - Prob. 9TFQ Ch. 8.2 - Prob. 10TFQ Ch. 8.2 - Prob. 1E Ch. 8.2 - Prob. 2E Ch. 8.2 - Prob. 3E Ch. 8.2 - 4. Find an algorithm for finding the smallest...Ch. 8.2 - Prob. 5E Ch. 8.2 - 6. (a) [BB] Justify the statement made in...Ch. 8.2 - Prob. 7E Ch. 8.2 - Prob. 8E Ch. 8.2 - Prob. 9E Ch. 8.2 - Prob. 10E Ch. 8.2 - Prob. 11E Ch. 8.2 - Prob. 12E Ch. 8.2 - Prob. 13E Ch. 8.2 - Prob. 14E Ch. 8.2 - Prob. 15E Ch. 8.2 - Prob. 16E Ch. 8.2 - Prob. 17E Ch. 8.2 - Prob. 18E Ch. 8.2 - Prob. 19E Ch. 8.2 - Prob. 20E Ch. 8.2 - Prob. 21E Ch. 8.2 - Prob. 22E Ch. 8.2 - Prob. 23E Ch. 8.2 - Prob. 24E Ch. 8.2 - Prob. 25E Ch. 8.2 - The Russian peasant method is used to multiply two...Ch. 8.2 - Prob. 27E Ch. 8.2 - Prob. 28E Ch. 8.3 - Prob. 1TFQ Ch. 8.3 - Prob. 2TFQ Ch. 8.3 - (Answers can be found in the back of the book.)...Ch. 8.3 - Prob. 4TFQ Ch. 8.3 - Prob. 5TFQ Ch. 8.3 - (Answers can be found in the back of the book.) 6....Ch. 8.3 - Prob. 7TFQ Ch. 8.3 - Prob. 8TFQ Ch. 8.3 - Prob. 9TFQ Ch. 8.3 - Prob. 10TFQ Ch. 8.3 - Prob. 1E Ch. 8.3 - Prob. 2E Ch. 8.3 - Describe a ternary search algorithm, which...Ch. 8.3 - Prob. 4E Ch. 8.3 - Prob. 5E Ch. 8.3 - Prob. 6E Ch. 8.3 - Prob. 7E Ch. 8.3 - Prob. 8E Ch. 8.3 - Prob. 9E Ch. 8.3 - Prob. 10E Ch. 8.3 - Prob. 11E Ch. 8.3 - Prob. 12E Ch. 8.3 - Prob. 13E Ch. 8.3 - Prob. 14E Ch. 8.3 - Prob. 15E Ch. 8.3 - Prob. 16E Ch. 8.3 - Prob. 17E Ch. 8.3 - [BB] Show the steps involved in the application of...Ch. 8.3 - Prob. 19E Ch. 8.3 - The Binary search Algorithm we have presented...Ch. 8.3 - Prob. 21E Ch. 8.3 - Prob. 22E Ch. 8.3 - Prob. 23E Ch. 8.3 - Prob. 24E Ch. 8.3 - Prob. 25E Ch. 8.3 - Prob. 26E Ch. 8.4 - (Answers can be found in the back of the book.) 1....Ch. 8.4 - Prob. 2TFQ Ch. 8.4 - Prob. 3TFQ Ch. 8.4 - Prob. 4TFQ Ch. 8.4 - Prob. 5TFQ Ch. 8.4 - Prob. 6TFQ Ch. 8.4 - Prob. 7TFQ Ch. 8.4 - Prob. 8TFQ Ch. 8.4 - Prob. 9TFQ Ch. 8.4 - Prob. 10TFQ Ch. 8.4 - Prob. 1E Ch. 8.4 - Use the procedure outlined in this section to list...Ch. 8.4 - Prob. 3E Ch. 8.4 - Prob. 4E Ch. 8.4 - Prob. 5E Ch. 8.4 - Prob. 6E Ch. 8.4 - Prob. 7E Ch. 8.4 - 8. (a) List, in the lexicographic order, the...Ch. 8.4 - Prob. 9E Ch. 8.4 - Prob. 10E Ch. 8.4 - Prob. 11E Ch. 8.4 - Prob. 12E Ch. 8 - Describe how Horners Algorithm evaluates f(x) when...Ch. 8 - Prob. 2RE Ch. 8 - 3. Let be an integer, let , and let be a subset of...Ch. 8 - Suppose we want an algorithm that, for an input of...Ch. 8 - Prob. 5RE Ch. 8 - Prob. 6RE Ch. 8 - Prob. 7RE Ch. 8 - Prob. 8RE Ch. 8 - (Requires a little knowledge of calculus) Show...Ch. 8 - Prob. 10RE Ch. 8 - Prob. 11RE Ch. 8 - 12. Sort the list 9,-3,1,0,-4,5,3 into increasing...Ch. 8 - 13. In the lexicographic ordering of all...Ch. 8 - Prob. 14RE

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To prove: 3 n 3 − 5 n 2 + 2 n + 1 © n 3

Solution Summary: The author states that the triangle inequality holds left|f(n)|a+bright|le

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To prove: 3n3−5n2+2n+1©n3

To prove: n4−2n3+3n2−5n+7©n4

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Chapter 8 Solutions