Math  /  Algebra

QuestionExercise 1. Express the following in Big- Ω\Omega, Big-O, or Big-Theta notation as appropriate. (a) n23n(n2)4n2n^{2} \leq 3 n(n-2) \leq 4 n^{2}, for every integer n3n \geq 3. (b) 12n2n(3n2)2\frac{1}{2} n^{2} \leq \frac{n(3 n-2)}{2}, for every integer n3n \geq 3 (c) 0n(3n2)2n20 \leq \frac{n(3 n-2)}{2} \leq n^{2}, for every integer n2n \geq 2.

Studdy Solution

STEP 1

1. We are dealing with asymptotic notation: Big-O, Big-Ω, and Big-Θ.
2. We need to analyze the inequalities given for large n n .
3. We will express the growth rates of the functions in terms of n n .

STEP 2

1. Analyze inequality (a) and express it in asymptotic notation.
2. Analyze inequality (b) and express it in asymptotic notation.
3. Analyze inequality (c) and express it in asymptotic notation.

STEP 3

For inequality (a), we have:
n23n(n2)4n2 n^{2} \leq 3n(n-2) \leq 4n^{2}
First, simplify the middle term:
3n(n2)=3n26n 3n(n-2) = 3n^2 - 6n
Compare this with the bounds:
- The lower bound is n2 n^2 , which is O(n2) O(n^2) . - The upper bound is 4n2 4n^2 , which is O(n2) O(n^2) .
Since 3n26n 3n^2 - 6n is sandwiched between two quadratic functions, it is Θ(n2) \Theta(n^2) .

STEP 4

For inequality (b), we have:
12n2n(3n2)2 \frac{1}{2}n^{2} \leq \frac{n(3n-2)}{2}
Simplify the right-hand side:
n(3n2)2=3n22n2 \frac{n(3n-2)}{2} = \frac{3n^2 - 2n}{2}
Compare this with the lower bound:
- The lower bound is 12n2 \frac{1}{2}n^2 , which is Ω(n2) \Omega(n^2) .
Since 3n22n2 \frac{3n^2 - 2n}{2} is greater than or equal to a quadratic function for n3 n \geq 3 , it is Ω(n2) \Omega(n^2) .

STEP 5

For inequality (c), we have:
0n(3n2)2n2 0 \leq \frac{n(3n-2)}{2} \leq n^{2}
Simplify the middle term:
n(3n2)2=3n22n2 \frac{n(3n-2)}{2} = \frac{3n^2 - 2n}{2}
Compare this with the bounds:
- The upper bound is n2 n^2 , which is O(n2) O(n^2) .
Since 3n22n2 \frac{3n^2 - 2n}{2} is less than or equal to a quadratic function, it is O(n2) O(n^2) .
The expressions in asymptotic notation are: (a) Θ(n2) \Theta(n^2) (b) Ω(n2) \Omega(n^2) (c) O(n2) O(n^2)

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