Question

Show the Dirichlet Product is commutative & associative

Answer 1

Just show that the order in which you do it doesn't matter by doing it both ways.

Answer 2

i wish it was that simple :(

Answer 3

What do you mean? Just show the convolution of f*g is the same as g*f for associative yeah?

Answer 4

f * g = \[\sum_{d|n}^{}f(d)g(n/d)\]

g*f = \[\sum_{d|n}^{}g(d)f(n/d)\]

where do i go from here?

Answer 5

i can write
f * g =
\[\sum_{de=n}^{}f(d)g(e)\]
and
g*f =
\[\sum_{de=n}^{}g(d)f(e)\]

Answer 6

see if u can use this,

\(\large \sum \limits_{d|n}^{}f(d)g(n/d) = \sum \limits_{\frac{n}{d}|n}^{}f(n/d)g(d) \)

if d is a divisor, then n/d is also a divosor. so both sums are essentially same

Answer 7

oh i see now, would i be require to show that n/d is a divior or is that just a definition?

Answer 8

we dont have to prove it i think its self evident :

n = d * n/d

Answer 9

whenever d is a divisor/factor, n/d is also a divisor
both have 1-1 mapping

Answer 10

oh i see alright thanks alot for your help! i can figure out associativity now thanks

Answer 11

np :)