Michael Guantonio
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 Nov 21 awarded Notable Question Sep 24 awarded Autobiographer May 16 awarded Popular Question Sep 15 comment Proving a factorial is not a certain complexity I would believe that the series would only increase by one. Since the limit itself is the node going to infinity there is really no change for n+1. n+1 will still basically be a very large number as it approaches infinity. Sep 15 awarded Scholar Sep 15 comment Calculating Average Case Complexity That was a good answer and very easy to follow. I feel that I can now solve a case complexity problem for a series that has several conditions. Basically it is a matter of defining the problem space. Matching conditions and then making a final deductive step. Sep 15 accepted Calculating Average Case Complexity Sep 15 awarded Supporter Sep 15 comment Proving a factorial is not a certain complexity @Steven Stadnicki Do you have any good resources that can help me remove my confusion on this? I must say that my algorithms teacher is not making it very clear and my discrete teacher decided to gloss over the subject as well. So bascially in the CS side of my classes, we all just put the two together so it is hard to separate. Sep 14 asked Proving a factorial is not a certain complexity Sep 14 comment Algorithmic Complexity of $i^2$ How can I deal with a more general function for this problem. Like proving for the general case that $$\sum_{i=1}^n i^k is O(n^{k-1})$$ Sep 14 asked Prove a formula is corect Sep 14 asked Calculating Average Case Complexity Sep 11 awarded Commentator Sep 11 comment Algorithmic Complexity of $i^2$ Still not quite sure how you got this. I can see that you took the general i case and said that it was less than or equal to the n max case. But how you determined because of that the complexity is $O(n^{3})$ baffles me. Sep 11 comment Algorithmic Complexity of $i^2$ @StevenStadnicki Sorry, the computer science student in me is trying to make since of it all. This means creating relations that are not there. Sep 11 awarded Editor Sep 11 comment If $f(n) = \sum_{i = 0}^n X_{i}$, then show by induction that $f(n) = f(n - 1) + X_{n-1}$ I have mistakenly given you the wrong equation. Sep 11 comment If $f(n) = \sum_{i = 0}^n X_{i}$, then show by induction that $f(n) = f(n - 1) + X_{n-1}$ I had mistakenly given you the wrong equation. Sep 11 revised If $f(n) = \sum_{i = 0}^n X_{i}$, then show by induction that $f(n) = f(n - 1) + X_{n-1}$ added 232 characters in body