Moebius Transform Sum: Understanding the mu(x) Function for Prime Numbers

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let be the sum (over all the divisors d of n):

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

f(p)= \mu (p)g(1) + \mu (1) g(p) is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p. :rolleyes:
 
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lokofer said:
let be the sum (over all the divisors d of n):

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

f(p)= \mu (p)g(1) + \mu (1) g(p) 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. :rolleyes:

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

Look at the definition of the mobius function.

Complete this step and \mu(1) and \mu(p) will be apparant.
 

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