MObius transform sum

  • Thread starter lokofer
  • Start date
  • #1
106
0
let be the sum (over all the divisors d of n):

[tex] f(n)= \sum_{d|n} \mu (n/d)g(d) [/tex] my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

[tex] f(p)= \mu (p)g(1) + \mu (1) g(p) [/tex] is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p. :uhh:
 

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
1,992
1
lokofer said:
let be the sum (over all the divisors d of n):

[tex] f(n)= \sum_{d|n} \mu (n/d)g(d) [/tex] my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

[tex] f(p)= \mu (p)g(1) + \mu (1) g(p) [/tex] is that correct?

Correct.

lokofer said:
...now the question is to know what's the value of mu(x) function for x=1 or p. :uhh:

Step #1 when trying to learn about mobius inversion and such:

Look at the definition of the mobius function.

Complete this step and [tex]\mu(1)[/tex] and [tex]\mu(p)[/tex] will be apparant.
 

Related Threads on MObius transform sum

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
7
Views
7K
Top