Feb 15, 2014 · I got confused when comparing 2n with 22n. Here , f(n)=2n ...
Missing: https:// | Show results with:https://
Apr 20, 2015 · basically the idea is that as n grows, both the functions 'n^2` and your function, behave similarly, ther'es a constant diffidence in their ...
Missing: https:// 21675/ 22n
People also ask
Is the following statement true: 22n O 2n?
Is 2n 1 O 2n is 22n O 2n?
Is 2n the same as n in Big O?
What is the most efficient Big O?
Jan 31, 2011 · First case is obviously true - you just multiply the constant C in by 2. Current answers to the second part of the question, ...
Mar 18, 2015 · And because it equals 2 according to what I've found on the internet, then f(n) and g(n) have an equal growth, therefore f(n) is part of O(n2).
Oct 29, 2018 · My answer is O(n^2) because it's the term with the highest degree. However, I'm not really sure how to show it. Am I right by saying that it has ...
Aug 18, 2022 · It's just equal to O(n). You can have O(f(n)) for any f, including f(n) = 2n. The definition of Big-O permits all sorts of useless ones like O( ...
May 19, 2011 · I am aware that with the definition of Big-O if an algorithm is O(n^2), then it is also O(n!), so to be more precise I am really looking for the ...
Duration: 18:35
Posted: Oct 7, 2023
Posted: Oct 7, 2023
In order to show you the most relevant results, we have omitted some entries very similar to the 8 already displayed.
If you like, you can repeat the search with the omitted results included. |