This was extremely interesting. For some more examples, in the previous lecture you mentioned some additive functions like omega(n) = # of distinct prime divisors of n, and you pointed out we could exponentiate them to get multiplicative functions. I took the Dirichlet series of b^omega(n) and ended up with an interesting looking identity: b^omega(n) = sum over square-free d|n of (b-1)^omega(d). Once you see it it's not hard to confirm by elementary means. I think I'll take a break then look at b^Omega(n).
@James-kg8fp4 жыл бұрын
I think you are tacitly using that no two distinct arithmetic functions have the same Dirichlet series. For those wondering how to prove this, consider the limit as s goes to infinity of k^s(F(s)-G(s)) where k is the first index that f and g differ.
@omargaber31224 жыл бұрын
Excuse me I dont understand what you say Do you mean there is an error in the explanation in the video? Or is this just an additional clarification?
@James-kg8fp4 жыл бұрын
@@omargaber3122 Not an error! More of an omission I was filling in. He uses relationships amongst Dirichlet series F,G and then concludes things about the underlying arithmetic functions f,g. This is a correct and useful thing to do, but it is only possible because if two arithmetic functions have the same Dirichlet series then they are the same function, a fact which was not mentioned in the video.
@omargaber31224 жыл бұрын
1-Do you mean that this only applies to functions that have the same arithmetic functions? 2-Could you PLEASE cite an example to the contrary? 3-Do you understand British Dyer's cojecture?
@richarde.borcherds79984 жыл бұрын
Alternatively you can just work with formal Dirichlet series, where convergence does not matter, and the Dirichlet series is just a suggestive way of writing an infinite sequence of integers.
@James-kg8fp4 жыл бұрын
@@richarde.borcherds7998 I like the idea of using formal series to avoid issues of convergence, which also avoids the problem of two different functions having the same Dirichlet series, since obsiously their formal series would be different. However, for formal series it seems one would have to justify why operations like F(s)/G(s) * G(s) = F(s) make sense. After all, what does division or multiplication of formal series mean without appealing to evaluations as series of reals? In that case one would have to do the bookkeeping of showing that the calculus of formal series is consistent.
@dneary3 жыл бұрын
I wonder if the Dirichlet series plus Mobius inversion formula gives a way to calculate a closed form for the sum of floor(n/i) for i=1..n - it seems like there is a close relationship between the floor and divisors of n - I have figured out that \sum_{i=1}^n [n/i] = \sum_{i=1}^n \sigma_0(i) - is there a nice way to get a closed form for this?
@martinepstein98263 жыл бұрын
In analytic number theory does one take analytic continuations of all these Dirichlet series and subject them to complex analysis?
@cboniefbr4 жыл бұрын
I love the you explain things, it all seems very fluid! Thank you so much for your content!
@martinepstein98264 жыл бұрын
Now that you've covered the Moebius function a video on Mertens conjecture could be interesting.
@omargaber31224 жыл бұрын
You are wonderfull professor.👏👏
@omargaber31224 жыл бұрын
Thank you very much doctor , please explain brich-dyre conjecture.