Question

I have this hunch about the behavior of the Dirichlet convolution of two types of functions, can anyone help me?

Answer 1

So specifically the functions I'm using are: \[\large u(n)=1 \\ \large N(n)=n \\ \large \tau(n)=\sum_{d | n} 1 \\ \large \sigma(n) = \sum_{d|n}d\] and I realized their dirichlet convolutions are linked in a kind of intuitive way: \[\large \sigma * u = \tau * N\] but I see that we could generalize these to become: \[ \large N_k(n)=n^k \\ \large \sigma_k(n) = \sum_{d|n}d^k\]\[\large \sigma_1 * N_0 = \sigma_0 * N_1\]

So I guess what I was wondering is are there other relations like this, such as \[\large \sigma_0*N_3 =\sigma_2 *N_1=\sigma_1*N_2=\sigma_0*N_3\]

Answer 2

Actually I think I figured it out, since I know we can write: \[ \large \sigma_k = \sum_{d|n}N_k=N_k*N_0\] Then I guess as we're pretty much allowed to do division and write: \[\large \frac{\sigma_k}{N_k}=N_0\] and we can say this is true for any k or other value too so \[\large \frac{\sigma_k}{N_k}=\frac{\sigma_l}{N_l} \\ \large N_l * \sigma_k = \sigma_l *N_k\] Interesting, even if what I did isn't exactly allowed I think it'll still work intuitively, and in a way we sort of have a whole set numbers here, which seems kind of interesting to play around with.

Answer 3

I guess a better proof would be to do this fact that the Dirichlet convolution is commutative \[\large \sigma_k *\sigma_l = \sigma_l * \sigma_k \] use the fact that by definition \[\large \sigma_k=N_k*N_0\]\[\large \sigma_k *N_l*N_0 = \sigma_l * N_k*N_0\] remembering that \[\large N_0=u\] and that \[\large u* \mu = I\] we can convolve both sides of the equation with the mobius function to get the identity, done!

Answer 4

@rational I guess all this is extra to prove it, but now I was wondering if there was any significance to \[\large \tau*N = \sigma * u\] and the generalization here or what cause this seems interesting but I'm pretty new to this so I don't know.

Answer 5

You are way ahead of majority of openstudy users...

Answer 6

Are you sure you need help?