came up with this cute lil guy in the comments of a previous thread.. Thought it deserved its own tweetpic.twitter.com/tCL4x65MxY
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In fact, the general form isn't difficult to prove at all.. Try it yourself!pic.twitter.com/PcV24x4B71
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Asking myself and the audience: If this integral identity implies a generalization of good old Nicomachus? Where the summation goes from 1 to p: (The case n=1, k=1 and n=2, k=1 are trivial and the case n=1, k=2 yields the identity of Nicomachus. What about k>1? k, n natural).pic.twitter.com/Uf5RcojWPH
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I was thinking that exact same thing!!! let me know if you come up with something!
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How are you getting an (n+1)^{k} for the r.h.s. integral?pic.twitter.com/pi6XxUKC24
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You’re forgetting the bounds
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That last exponential is (n+1)^k -1 or (n+1)^(k-1)??
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You can figure out by using the n=1 case;)
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Thanks. This actually might be useful with regard to a theorem i am trying to prove about the mathematical characteristics of species abundance distributions.
Thanks. Twitter will use this to make your timeline better. UndoUndo
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Thanks. Twitter will use this to make your timeline better. UndoUndo
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