Explanation Given information: c is any positive real number such that for every n ≥ 1 , f ( n ) = c Formula used: Let f and g be real valued functions defined on the same nonnegative integers, with g ( n ) ≥ 0 for every integer n ≥ r , where r is positive real number. Then, f is of order g , written f ( n ) is Θ ( g ( n ) ) , if and only if, there exist positive real numbers A , B and k ≥ r such that A g ( n ) ≤ f ( n ) ≤ B g ( n ) for every integer k ≥ a . Proof: Let A and B be any positive real numbers such that A ≤ c and B ≥ c

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ISBN: 9780357540244

Author: EPP

Publisher: CENGAGE L

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Chapter 11.2, Problem 41ES

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To prove:

That f(n) is Θ(1) for every n≥1, where f(n)=c and c is a positive real number.

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Chapter 11.2, Problem 40ES

Chapter 11.2, Problem 42ES

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To prove: That f ( n ) is Θ ( 1 ) for every n ≥ 1 , where f ( n ) = c and c is a positive real number.

Solution Summary: The author proves that f(n)is Theta (1 ) is a positive real number.

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To prove:

That f(n) is Θ(1) for every n≥1, where f(n)=c and c is a positive real number.

Want to see the full answer?

Chapter 11 Solutions

To prove: That f ( n ) is Θ ( 1 ) for every n ≥ 1 , where f ( n ) = c and c is a positive real number.

Solution Summary: The author proves that f(n)is Theta (1 ) is a positive real number.

Concept explainers

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To prove: That f(n) is Θ(1) for every n≥1, where f(n)=c and c is a positive real number.

Want to see the full answer?

Chapter 11 Solutions

To prove:

That f(n) is Θ(1) for every n≥1, where f(n)=c and c is a positive real number.