RK4 Posted September 19, 2005 Posted September 19, 2005 Hi all! I'm supposed to prove the following: Suppose that m is a positive real number. Show that Sigma(j^m), j runs from 1 to n is O(n^(m+1)). So we have: Sigma(j^m), j runs from 1 to n = 1^m + 2^m + 3^m + . . . + n^m I think we should use induction on m here to prove this. Basis Step: m = 1 This is definitely true for m = 1 because we have Sigma(j^1), j runs from 1 to n = n(n+1)/2 which is O(n^2) Now, O(n^2) = O(n^(1+1) = O(n^(m+1) Inductive Step: Assume Sigma(j^m), j runs from 1 to n is O(n^(m+1)) to be true. That is, it is true for m = n. (Inductive Hypothesis) Then must show it is also true for m = n + 1: Sigma(j^m), j runs from 1 to n + 1 = Sigma(j^m), j runs from 1 to n + j^(n+1) = O(n^(m+1)) + j^(n+1) (By Inductive Hypothesis) = . . . At this point how do I conclude that the whole thing is O(n^(m+1))? Any help will be appreciated. I'm trying to get a hold of these big O proofs.
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