Answer:
1. n! is O(n^n).
2. 1+2+...+n is O(n^2).
3. 8n + log₂(n) is O(n).
4. n²(7n - 2) is O(n³).
5. nlog₂(n) is O(n²), but it is not Θ(n²).
6. n¹⁰⁰ is O(2ⁿ), but it is not Θ(2ⁿ).
7. Θ(5n² + 4n + 2) = Θ(n² + 100n).
8. Θ(log₂(n)) = Θ(log₅(6n)).
Step-by-step explanation:
To analyze the asymptotic complexity of these functions, we will use the Big O notation (O) and the Theta notation (Θ). Let's examine each statement one by one:
1. n! is O(n^n):
To show that n! is O(n^n), we need to find a positive constant c and a positive integer n₀ such that for all n ≥ n₀, n! ≤ c * (n^n).
By Stirling's approximation, n! ≈ √(2πn) * (n/e)^n.
Let's choose c = e and n₀ = 1. For all n ≥ n₀:
n! ≈ √(2πn) * (n/e)^n ≤ e * n^n
Thus, n! is O(n^n).
2. 1+2+...+n is O(n^2):
The sum of the first n natural numbers, 1+2+...+n, is equal to n(n+1)/2.
We want to find a positive constant c and a positive integer n₀ such that for all n ≥ n₀, 1+2+...+n ≤ c * n^2.
Let's choose c = 1/2 and n₀ = 1. For all n ≥ n₀:
1+2+...+n = n(n+1)/2 ≤ (n^2)/2
Thus, 1+2+...+n is O(n^2).
3. 8n + log₂(n) is O(n):
To show that 8n + log₂(n) is O(n), we need to find a positive constant c and a positive integer n₀ such that for all n ≥ n₀, 8n + log₂(n) ≤ c * n.
Let's choose c = 9 (greater than the maximum coefficient) and n₀ = 1. For all n ≥ n₀:
8n + log₂(n) ≤ 9n
Thus, 8n + log₂(n) is O(n).
4. n²(7n - 2) is O(n³):
We want to show that n²(7n - 2) is O(n³), which means we need to find a positive constant c and a positive integer n₀ such that for all n ≥ n₀, n²(7n - 2) ≤ c * n³.
Let's choose c = 8 and n₀ = 1. For all n ≥ n₀:
n²(7n - 2) ≤ 8n³
Thus, n²(7n - 2) is O(n³).
5. nlog₂(n) is O(n²), but Θ(nlog₂(n)) ≠ Θ(n²):
nlog₂(n) is indeed O(n²) since for sufficiently large n, nlog₂(n) ≤ cn², where c = 1. However, it is not Θ(n²) because it does not satisfy the second condition of Θ notation, which requires a lower bound as well.
6. n¹⁰⁰ is O(2ⁿ), but Θ(n¹⁰⁰) ≠ Θ(2ⁿ):
n¹⁰⁰ is indeed O(2ⁿ) since for sufficiently large n, n¹⁰⁰ ≤ c * 2ⁿ, where c is a large constant. However, it is not Θ(2ⁿ) because it grows significantly faster than 2ⁿ.
7. Θ(5n² + 4n + 2) = Θ(n² + 100n):
To show that Θ(5n² + 4n + 2) = Θ(n² + 100n), we need to prove that both functions have the same upper and lower bounds.
Let's start with the upper bound:
5n² + 4n + 2 ≤ 5n² + 100n for all n ≥ 1
Thus, Θ(5n² + 4n + 2) is O(n² + 100n).
Now, let's move to the lower bound:
5n² + 4n + 2 ≥ n² + 100n for all n ≥ 1
Thus, Θ(5n² + 4n + 2) is Ω(n² + 100n).
Since Θ(5n² + 4n + 2) is both O(n² + 100n) and Ω(n² + 100n), we can conclude that Θ(5n² + 4n + 2) = Θ(n² + 100n).
8. Θ(log₂(n)) = Θ(log₅(6n)):
To show that Θ(log₂(n)) = Θ(log₅(6n)), we need to prove that both functions have the same upper and lower bounds.
Let's start with the upper bound:
log₂(n) ≤ log₅(6n) for all n ≥ 1
Thus, Θ(log₂(n)) is O(log₅(6n)).
Now, let's move to the lower bound:
log₂(n) ≥ log₅(6n) for all n ≥ 1
Thus, Θ(log₂(n)) is Ω(log₅(6n)).
Since Θ(log₂(n)) is both O(log₅(6n)) and Ω(log₅(6n)), we can conclude that Θ(log₂(n)) = Θ(log₅(6n)).
In summary:
1. n! is O(n^n).
2. 1+2+...+n is O(n^2).
3. 8n + log₂(n) is O(n).
4. n²(7n - 2) is O(n³).
5. nlog₂(n) is O(n²), but it is not Θ(n²).
6. n¹⁰⁰ is O(2ⁿ), but it is not Θ(2ⁿ).
7. Θ(5n² + 4n + 2) = Θ(n² + 100n).
8. Θ(log₂(n)) = Θ(log₅(6n)).