WebComputer Science. Computer Science questions and answers. Are each of the following true or false? (a) 3 n^2 + 10 n log n = O (n log n) (b) 3 n^2 + 10 n log n = Omega (n^2) (c) 3 n^2 + 10 n log n = Theta (n^2) (d) n log n + n/2 = O (n) (e) 10 SQRT (n) + log n = O (n) (f) SQRT (n) + log n = O (log n) (g) SQRT (n) + log n = Theta (log n) (h) SQRT ... WebJul 31, 2024 · $\begingroup$ "Big O" is time complexity that describes the worst case scenario.. so, you want to look for the term that will produce the highest values when considering values of n while approaching infinity. As for the other two terms, they will "fall to the side", or really, become so small in contrast to the overall resulting value that the …
Is $\\log(n!) \\in\\Theta(n \\log n)$ - Mathematics Stack Exchange
WebSee: Logarithm rules Logarithm product rule. The logarithm of the multiplication of x and y is the sum of logarithm of x and logarithm of y. log b (x ∙ y) = log b (x) + log b (y). For example: log 10 (3 ∙ 7) = log 10 (3) + log … WebJun 10, 2015 · T(n) = sqrt(n) * T(sqrt(n)) + n Given solution is O(log log n). But my solution is O(n log log n). 'wolframalpha'' shows the answer same as mine. You can find the solution here. Can anyone confirm the solution and provide an explantion? current beenleigh temperature
n*log n and n/log n against polynomial running time
WebFeb 23, 2024 · log n^2 is equivalent to 2logn which grows at the same rate as logn, as I disregard the factors and constants. but if I was to square the whole term so that I end up … WebIn fact, you can replace all the 2 s in the above argument with any k > 1, and obtain that log(n!) is eventually greater than Cnlogn for any C < 1 / k. Letting k approach 1, you can conclude that in fact log(n!) ∼ nlogn. "For n large, log2 / 2 ≪ logn / 2 and log(n / 2) ≪ nlogn, so we get log(n!) ≥ Cnlogn for any C < 1 / 2 for ... WebSep 26, 2015 · More precisely, if there are $\Theta(n)$ terms that are all $\Theta(\log n)$ in size, then their sum will indeed be $\Theta(n \log n)$ and we can conclude $\log n! \in \Omega(n \log n)$. Taking half of the terms is merely the simplest idea to describe and calculate, and fortunately it satisfies the needed conditions. current beef prices on the hoof