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The following statement is valid. log(n!) = θ(n log n).
- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The following statement is valid. log(n!) =θ(n log n).a)Trueb)Fal...
Order of growth of [Tex]log n![/Tex] and [Tex]nlog n[/Tex] is same for large values of [Tex]n[/Tex], i.e., [Tex] heta (log n!) = heta (nlog n)[/Tex]. So time complexity of fun() is [Tex] heta (nlog n)[/Tex]. The expression [Tex] heta (log n!) = heta (nlog n)[/Tex] can be easily derived from following Stirling's approximation (or Stirling's formula). [Tex] log n! = nlog n - n +O(log(n)) [/Tex]
Most Upvoted Answer
The following statement is valid. log(n!) =θ(n log n).a)Trueb)Fal...
Explanation:
To prove the given statement, we can use Stirling's approximation formula which states that:
n! ≈ √(2πn)(n/e)^n
Taking the logarithm of both sides, we get:
log(n!) ≈ log(√(2πn)(n/e)^n)
log(n!) ≈ log(2πn)/2 + n log(n/e)
log(n!) ≈ log(2πn)/2 + n(log(n) - log(e))
log(n!) ≈ log(2πn)/2 + n(log(n) - 1)
Now, we can simplify the above equation as follows:
log(n!) ≈ n(log(n) - 1) (since log(2πn)/2 is a constant)
Therefore, we can conclude that:
log(n!) = O(n log n)
Hence, the given statement is true.
HTML:
Explanation:
To prove the given statement, we can use Stirling's approximation formula which states that:
n! ≈ √(2πn)(n/e)^n
Taking the logarithm of both sides, we get:
log(n!) ≈ log(√(2πn)(n/e)^n)
log(n!) ≈ log(2πn)/2 + n log(n/e)
log(n!) ≈ log(2πn)/2 + n(log(n) - log(e))
log(n!) ≈ log(2πn)/2 + n(log(n) - 1)
Now, we can simplify the above equation as follows:
log(n!) ≈ n(log(n) - 1) (since log(2πn)/2 is a constant)
Therefore, we can conclude that:
log(n!) = O(n log n)
Hence, the given statement is true.
HTML:
Explanation:
- To prove the given statement, we can use Stirling's approximation formula which states that:
- n! ≈ √(2πn)(n/e)^n
- Taking the logarithm of both sides, we get:
- log(n!) ≈ log(√(2πn)(n/e)^n)
- log(n!) ≈ log(2πn)/2 + n log(n/e)
- log(n!) ≈ log(2πn)/2 + n(log(n) - log(e))
- log(n!) ≈ log(2πn)/2 + n(log(n) - 1)
- Now, we can simplify the above equation as follows:
- log(n!) ≈ n(log(n) - 1) (since log(2πn)/2 is a constant)
- Therefore, we can conclude that:
- log(n!) = O(n log n)
- Hence, the given statement is true.
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